calculus one and several variables 10e salas solutions manual ch12

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU

JWDD027-12 JWDD027-Salas-v1 December 2, 2006 16:55

632 SECTION 12.1

CHAPTER 12

SECTION 12.1

1. 1 + 4 + 7 = 12 2. 2 + 5 + 8 + 11 = 26

3. 1 + 2 + 4 + 8 = 15 4. 12 + 1

4 + 18 + 1

16 = 1516

5. 1 − 2 + 4 − 8 = −5 6. 2 − 4 + 8 − 16 = −10

7. 13 + 1

9 + 127 = 13

27 8. − 16 + 1

24 − 1120 = − 2

15

9. 1 + 14 + 1

16 + 164 = 85

64 10. 1 − 14 + 1

16 − 164 = 51

64

11.11∑

n=1

(2n− 1) 12.10∑k=1

(−1)k+1(2k − 1) 13.35∑k=1

k(k + 1)

14.n∑

k=1

mk Δxk 15.n∑

k=1

MkΔxk 16.n∑

k=1

f(x∗k) Δxk

17.10∑k=3

12k

,

7∑i=0

12i+3

18.10∑k=3

kk

k!,

7∑i=0

(i + 3)i+3

(i + 3)!

19.10∑k=3

(−1)k+1 k

k + 1,

7∑i=0

(−1)ii + 3i + 4

20.10∑k=3

12k − 3

,

7∑i=0

12i + 3

21. Set k = n + 3. Then n = −1 when k = 2 and n = 7 when k = 10.10∑k=2

k

k2 + 1=

7∑n=−1

n + 3(n + 3)2 + 1

=7∑

n=−1

n + 3n2 + 6n + 10

22.12∑

n=2

(−1)n

n− 1=

11∑k=1

(−1)k+1

k + 1 − 1=

11∑k=1

(−1)k+1

k

23. Set k = n− 3. Then n = 7 when k = 4 and n = 28 when k = 25.25∑k=4

1k2 − 9

=28∑

n=7

1(n− 3)2 − 9

=28∑

n=7

1n2 − 6n

24.15∑k=0

32k

k!=

13∑n=−2

32(n+2)

(n + 2)!= 81

13∑n=−2

32n

(n + 2)!

25. 0.a1a2 · · · an =a1

10+

a2

102+ · · · + an

10n=

n∑k=1

ak10k



SECTION 12.2 633

26.n∑

k=1

1√k

=1√1

+1√2

+1√3

+ · · · + 1√n≥ 1√

n+

1√n

+1√n

+ · · · + 1√n

=n√n

=√n.

27.50∑k=0

14k

= 1.3333 · · · 28.50∑k=1

1k2

∼= 1.62513

29.50∑k=0

1k!

= 2.71828 · · · 30.50∑k=0

(23

)k

∼= 3

SECTION 12.2

1.12; sn =

12

[1

1 · 2 +1

2 · 3 + · · · + 1(n)(n + 1)

]

=12

[(1 − 1

2

)+(

12− 1

3

)+ · · · +

(1n− 1

n + 1

)]=

12

[1 − 1

n + 1

]→ 1

2

2.12;

∞∑k=3

1k2 − k

=∞∑

k=3

(1

k − 1− 1

k

)= lim

n→∞

(12− 1

n

)=

12

3.1118

; sn =1

1 · 4 +1

2 · 5 + · · · + 1n(n + 3)

=13

[(1 − 1

4

)+(

12− 1

5

)+ · · · +

(1n− 1

n + 3

)]

=13

[1 +

12

+13− 1

n + 1− 1

n + 2− 1

n + 3

]→ 1

3

(1 +

12

+13

)=

1118

4.34;

∞∑k=0

1(k + 1)(k + 3)

=12

∞∑k=0

(1

k + 1− 1

k + 3

)=

12

limn→∞

(1 +

12− 1

n + 2− 1

n + 3

)=

34

5.103

;∞∑

k=0

310k

= 3∞∑

k=0

(110

)k

= 3(

11 − 1/10

)=

309

=103

6.56;

∞∑k=0

(−1)k

5k=

∞∑k=0

(−1

5

)k

=1

1 + 15

=56

7. −32;

∞∑k=0

1 − 2k

3k=

∞∑k=0

(13

)k

−∞∑

k=0

(23

)k

=1

1 − 1/3− 1

1 − 2/3=

32− 3 = −3

2

8.14;

∞∑k=0

12k+3

=18

∞∑k=0

12k

=18· 11 − 1

2

=14

9. 24; geometric series with a = 8 and r =23, sum =

a

1 − r= 24



634 SECTION 12.2

10.3

15, 616;

∞∑k=2

3k−1

43k+1=

∞∑k=0

3k+1

43k+7=

347

∞∑k=0

(343

)k

=347

· 11 − 3

43

=3

15, 616

11. Let x = 0.︷︸︸︷a1a2 · · · an

︷︸︸︷a1a2 · · · an · · · . Then

x =∞∑

k=1

a1a2 · · · an(10n)k

= a1a2 · · · an∞∑

k=1

(1

10n

)k

= a1a2 · · · an[

11 − 1/10n

− 1]

=a1a2 · · · an10n − 1

.

12. (a) Denote the partial sums of the first series by sn and those of the second series by tn and observe

that

sn = (a0 + a1 + · · · + aj) + tn. Obviously sn → L iff

tn = sn − (a0 + a1 + · · · + aj) → L− (a0 + a1 + · · · + aj).

Parts (b) and (c) follow from this equation.

13.1

1 + x=

11 − (−x)

=∞∑

k=0

(−x)k =∞∑

k=0

(−1)kxk

14.1

1 + x2=

11 − (−x2)

=∞∑

k=0

(−x2)k =∞∑

k=0

(−1)kx2k

15.x

1 − x= x

(1

1 − x

)= x

∞∑k=0

(xk) =∞∑

k=0

xk+1

16.x

1 + x= x · 1

1 − (−x)= x

∞∑k=0

(−x)k =∞∑

k=0

(−1)kxk+1

17.x

1 + x2= x

[1

1 − (−x2)

]= x

∞∑k=0

(−x2)k =∞∑

k=0

(−1)kx2k+1

18.x

1 + 4x2=

x

1 − (−4x2)= x

∞∑k=0

(−4x2)k =∞∑

k=0

(−1)k(2x)2k+1

19. 1 +32

+94

+278

+8116

+ · · · =∞∑

k=0

(32

)k

This is a geometric series with x = 32 > 1. Therefore the series diverges.

20. ak =14

(−54

)k

does not go to zero

21. limk→∞

(k + 1k

)k

= e �= 0



SECTION 12.2 635

22. ak =kk−2

3k=(k

3

)k

· 1k2

>2k

k2for k > 6, so ak → ∞

23. Rebounds to half its previous height:

s = 6 + 3 + 3 +32

+32

+34

+34

+ · · · = 6 + 6∞∑

k=0

12k

= 6 +6

1 − 12

= 18 ft.

24. s = 6 + 2h + 2h(h

6

)+ 2h

(h

6

)2

+ · · · = 6 + 2h∞∑

k=0

(h

6

)k

= 6 +11h

6 − h= 21

=⇒ 11h = 15(6 − h) =⇒ h = 4513

25. A principal x deposited now at r% interest compounded annually will grow in k years to

x(1 +

r

100

)k.

This means that in order to be able to withdraw nk dollars after k years one must place

nk

(1 +

r

100

)−k

dollars on deposit today. To extend this process in perpetuity as described in the text, the total deposit

must be∞∑

k=1

nk

(1 +

r

100

)−k

.

26. (a)∞∑

k=1

5000(

12

)k−1

(1.05)−k =50001.05

∞∑k=1

[1

2(1.05)

]k−1

=50001.05

· 11 − 1

2·1

∼= $9090.91

(b) 8000.81.06

1 − 0.81.06

∼= $2461.54

(c)N

1.05· 11 − 1

1.05

= 20N

27.∞∑

n=1

(910

)n

=

910

1 − 910

= 9 or $9

28. Total length removed =13

+29

+427

+ · · · =13

∞∑k=0

(23

)k

=13· 11 − 2

3

= 1

Some points: 0, 1, 13 ,

23 ,

19 ,

29 ,

79 ,

89 .

29. A = 42 + (2√

2)2 + 22 + (√

2)2 + 12 + · · · +[4(

1√2

)n]2+ · · ·

=∞∑

n=0

[4(

1√2

)n]2= 16

∞∑n=0

(12

)n

= 16 · 11 − 1

2

= 32



636 SECTION 12.2

30. (a) If∑

(ak + bk) converges, then∑

bk =∑

(ak + bk) −∑

ak would also converge.

(b) If ak = bk = 2k,∑

ak,∑

bk and∑

(ak + bk) diverge, but∑

(ak − bk) =∑

0 converges.

(c) If ak = 2k, bk = −2k,∑

ak,∑

bk and∑

(ak − bk) diverge, but∑

(ak + bk) =∑

0

converges.

31. Let L =∞∑

k=0

ak. Then

L =∞∑

k=0

ak =n∑

k=0

ak +∞∑

k=n+1

ak = sn + Rn.

Therefore, Rn = L− sn and since sn → L as n → ∞, it follows that Rn → 0 as n → ∞.

32. (a) By convergence, ak → 0, so1ak

diverges, hence∑ 1

akdiverges.

(b) If ak =√k, then

∑ak diverges and

∑ 1ak

diverges (Example 5)

If ak = 2k, then∑

ak diverges and∑ 1

ak=∑ 1

2kconverges.

33. sn =n∑

k=1

ln(k + 1k

)= [ ln(n + 1) − ln (n)] + [ lnn− ln(n− 1)] + · · · + [ ln 2 − ln 1] = ln(n + 1) → ∞

34. ak =(

k

k + 1

)k

=

⎛⎜⎝ 1

k + 1k

⎞⎟⎠

k

→ 1e�= 0

35. (a) sn =n∑

k=1

(dk − dk+1) = d1 − dn+1 → d1

(b) We use part (a).

(i)∞∑

k=1

√k + 1 −

√k√

k(k + 1)=

∞∑k=1

[1√k− 1√

k + 1

]= 1

(ii)∞∑

k=1

2k + 12k2(k + 1)2

=∞∑

k=1

12

[1k2

− 1(k + 1)2

]=

12

36. Use induction to verify the hint. Then

sn =1 − (n + 1)xn + nxn+1

(1 − x)2→ 1

(1 − x)2

Since −(n + 1)xn + nxn+1 → 0 for |x| < 1. This last statement follows from observing that nxn → 0.



SECTION 12.2 637

To see this, choose ε > 0 so that (1 + ε)|x| < 1. Since n1/n → 1, there exists k so that

n1/n < 1 + ε for n ≥ k.

Then for n ≥ k

|nxn| = |n1/nx|n ≤ ((1 + ε)|x|)n → 0

37. Rn =∞∑

k=n+1

14k

=

(14

)n+1

1 − 14

=1

3 · 4n ;

13 · 4n < 0.0001 =⇒ 4n > 3333.33 =⇒ n >

ln 3333.33ln 4

∼= 5.85

Take N = 6.

38. Rn =∞∑

k=n+1

(0.9)k = (0.9)n+1 11 − 0.9

= 9(0.9)n < 0.0001 =⇒ n ≥ ln(

0.00019

)ln 0.9

∼= 109

39. Rn =∞∑

k=n+1

1k(k + 2)

=12

∞∑k=n+1

(1k− 1

k + 2

)=

12

(1

n + 1+

1n + 2

);

12

(1

n + 1+

1n + 2

)< 0.0001 =⇒ n ≥ 9999. Take N = 9999.

40. Rn =∞∑

k=n+1

(23

)k

=(

23

)n+1 11 − 2

3

= 2(

23

)n

< 0.0001 =⇒ n ≥ ln(

0.00012

)ln(

23

) ∼= 25

41. |Rn| =∣∣∣∣ ∞∑k=n+1

xk

∣∣∣∣ =∣∣∣∣ xn+1

1 − x

∣∣∣∣ = |x|n+1

1 − x;

|x|n+1

1 − x< ε

|x|n+1 < ε(1 − x)

(n + 1) ln |x| < ln ε(1 − x)

n + 1 >ln ε(1 − x)

ln |x| [recall ln |x| < 0]

n >ln ε(1 − x)

ln |x| − 1

Take N to be smallest integer which is greater thanln ε(1 − x)

ln |x| .

42. sn = an − a1. Thus {sn} converges iff {an} converges.

Hence∞∑

k=1

(ak+1 − ak) converges iff {an} converges.



638 SECTION 12.3

SECTION 12.3

1. converges; basic comparison with∑ 1

k22. diverges; limit comparison with

∑ 1k


k24. diverges; basic comparison with

∑ 1k

5. diverges; basic comparison with∑ 1

k + 16. converges; basic comparison with

∑ 1k2

7. diverges; limit comparison with∑ 1

k8. converges; geometric with x = 2

5

9. converges; integral test,∫ ∞

1

tan−1 x

1 + x2dx = lim

b→∞

[12(tan−1 x)2

]b1

=3π2

32


k211. diverges; p-series with p = 2

3 ≤ 1


k313. diverges; divergence test, ( 3

4 )−k �→ 0

14. diverges; basic comparison with∑ 1

1 + 2k15. diverges; basic comparison with

∑ 1k


2

dx

x(lnx)2= lim

b→∞

[ −1lnx

]∞2

=1

ln 2

17. diverges; divergence test,1

2 + 3−k→ 1

2�= 0

18. converges; limit comparison with∑ 1

k4


k220. diverges; ak �→ 0.

21. diverges; integral test,∫ ∞

2

dx

x lnx= lim

b→∞[ ln (lnx)]b2 = ∞


2k23. converges; limit comparison with

∑ 2k

5k


k25. diverges; limit comparison with

∑ 1k

26. diverges; limit comparison with∑ 1√

k27. converges; limit comparison with

∑ 1k3/2


k



SECTION 12.3 639


1

xe−x2dx = lim

b→∞

[−1

2e−x2

]b1

=12e

30. converges; integral test:∫ ∞

1

x22−x3dx = lim

b→∞

[2−x3

−3 ln 2

]∞1

=1

6 ln 2


k2, 2 + sin k ≤ 3 for all k.

32. diverges; basic comparison with∑ 1√

k,

2 + cos k√k + 1

>1√k

33. Recall that 1 + 2 + 3 + · · · + k =k(k + 1)

2. Therefore

∑ 11 + 2 + 3 + · · · + k

=∑ 2

k(k + 1). This series converges; direct comparison with

∑ 2k2

34.∑ n

1 + 22 + 32 + · · · + n2=∑ n

16n(n + 1)(n + 2)

converges: limit comparison with∑ 1

n2


k2:∑ 2k

(2k)!=∑ 1

(2k − 1)(2k − 2) · · · 3 · 2 · 1 <∑ 1

k2


k2:∑ 2k!

(2k)!=∑ 1

k(2k − 1)(2k − 2) · · · (k + 1)<∑ 1

k2

37. Use the integral test:

Let u = lnx, du =1xdx :

∫1

x(lnx)pdx =

∫u−p du =

u1−p

1 − p+ C.

∫ ∞

1

1x(lnx)p

dx = limb→∞

∫ b

1

1x(lnx)p

dx = limb→∞

11 − p

(ln a)1−p

The series converges for p > 1.

38. If p ≤ 1,∑ ln k

kp>∑ 1

kpdiverges.

If p > 1, thenp− 1

2> 0, so for large k, ln k < k

p−12

Thenln k

kp<

kp−12

kp=

1

kp+12

. Sincep + 1

2> 1,

∑ 1

kp+12

converges

Hence so does∑ ln k

kp. so converges iff p > 1.

39. (a) Use the integral test:∫ ∞

0

e−αx dx = limb→∞

[− 1

αe−αx

]b0

=1α

converges.

(b) Use the integral test:∫ ∞

0

xe−αx dx = limb→∞

[− x

αe−αx− 1

α2e−αx

]b0

=1α2

converges.



640 SECTION 12.3

(c) The proof follows by induction using parts (a) and (b) and the reduction formula∫xneax dx =

xneax

a− n

a

∫xn−1eax dx [see Exercise 67, Section 8.2]

40.∫ ∞

n+1

dx

xp<

∞∑k=n+1

1kp

<

∫ ∞

n

dx

xp

=⇒ 1(p− 1)(n + 1)p−1

<

∞∑k=1

1kp

−n∑

k=1

1kp

<1

(p− 1)np−1

41. (a)4∑

k=1

1k3

∼= 1.1777 (b)1

2 · 52< R4 <

12 · 42

0.02 < R4 < 0.0313

(c) 1.1777 + 0.02 = 1.1977 <

∞∑k=1

1k3

< 1.1777 + 0.0313 = 1.2090

42. (a)4∑

k=1

1k4

= 1 +124

+134

+144

∼= 1.0788

(b)1

3(5)3< R4 <

13(4)3

=⇒ 0.0027 < R4 < 0.0052

(c) 1.0815 <

∞∑k=1

1k4

< 1.0840

43. (a) Put p = 2 and n = 100 in the estimates in Exercise 38. The result is:1

101< R100 <

1100

.

(b) Rn <1

(2 − 1)n2−1< 0.0001 =⇒ n > 10, 000 Take n = 10, 001.

44. (a)1

2(101)2< R100 <

12(100)2

=⇒ 0.000049 < R100 < 0.00005

(b) Rn <1

2n2< 0.0001 =⇒ n ≥ 71

(c)∞∑

k=1

1k3

∼=71∑k=1

1k3

∼= 1.20196

45. (a) Rn <1

(4 − 1)n4−1< 0.0001 =⇒ n3 > 3333 =⇒ n > 14.94 : Take n = 15.

(b) Rn <1

(4 − 1)n4−1< 0.001 =⇒ n3 > 333.33 =⇒ n > 6.93 : Take n = 7.

(c)∞∑

k=1

1k4

∼=7∑

k=1

1k4

∼= 1.082



SECTION 12.3 641

46. (a) Rn <1

4n4< 0.0001 =⇒ n ≥ 8

(b) Rn <1

(4)n4< 0.001 =⇒ n3 > 333.33 =⇒ n > 3.97 : Take n = 4.

(c)∞∑

k=1

1k5

∼=4∑

k=1

1k5

∼= 1.0363

47. (a) If ak/bk → 0, then ak/bk < 1 for all k ≥ K for some K. But then ak < bk for all k ≥ K

and, since∑

bk converges,∑

ak converges. [ The Basic Comparison Theorem 12.3.6.]

(b) Similar to (a) except that this time we appeal to part (ii) of Theorem 12.3.6.

(c)∑

ak =∑ 1

k2converges,

∑bk =

∑ 1k3/2

converges,1/k2

1/k3/2=

1√k→ 0

∑ak =

∑ 1k2

converges,∑

bk =∑ 1√

kdiverges,

1/k2

1/√k

=1

k3/2→ 0

(d)∑

bk =∑ 1√

kdiverges,

∑ak =

∑ 1k2

converges,1/k2

1/√k

=1

k3/2→ 0

∑bk =

∑ 1√k

diverges,∑

ak =∑ 1

kdiverges,

1/k1/√k

=1√k→ 0

48. (a) Since ak/bk → ∞, bk/ak → 0, so this follows from Exercise 45(b)

(b) Follows from Exercise 45(a)

(c)∑

ak =∑ 1√

kdiverges,

∑bk =

∑ 1k2

converges,1/√k

1/(k2)= k3/2 → ∞

∑ak =

∑ 1√k

diverges,∑

bk =∑ 1

kdiverges,

1/√k

1/k=

√k → ∞

(d)∑

bk =∑ 1

k2converges,

∑ak =

∑ 1k3/2

converges,1/k3/2

1/(k2)=

√k → ∞

∑bk =

∑ 1k2

converges,∑

ak =∑ 1√

kdiverges,

1/√k

1/(k2)= k3/2 → ∞

49. (a) Since∑

ak converges, ak → 0. Therefore there exists a positive integer N such that 0 < ak < 1

for k ≥ N. Thus, for k ≥ N, a2k < ak and so

∑a2k converges by the comparison test.

(b)∑

ak may either converge or diverge:∑

1/k4 and∑

1/k2 both converge;∑

1/k2 converges

and∑

1/k diverges.

50. Since 0 <

(ak − 1

k

)2

< ak2 +

1k2

,∑(

ak − 1k

)2

converges by comparison with

∑ak

2 +∑ 1

k2But

∑(ak − 1

k

)2

=∑

a2k − 2

∑ akk

+∑ 1

k2,

so∑ ak

kmust converge.



642 SECTION 12.4

51. 0 < L−n∑

k=1

f(k) = L− sn =∞∑

k=n+1

f(k) <∫ ∞

n

f(x) dx [see the proof of the integral test]

52. 0 < L− Sn <

∫ ∞

n

1x2 + 1

dx =π

2− arctan n < 0.001 =⇒ n > tan

(π2− 0.001

)∼= 1000

53. L− sn <

∫ ∞

n

xe−x2dx = lim

b→∞

∫ b

n

xe−x2dx

= limb→∞

[− 1

2e−x2

]bn

=12e−n2

12e−n2

< 0.001 =⇒ en2> 500 =⇒ n > 2.49; take N = 3.

54. (a) Set f(x) = 1/x in the proof of the integral test.

(b)n∑1

1k> 100 if ln(n + 1) > 100 =⇒ n + 1 > e100 ∼= 2.7 × 1043

55. Set f(x) = x1/4 − lnx. Then

f ′(x) =14x−3/4 − 1

x=

14x

(x1/4 − 4).

Since f(e12) = e3 − 12 > 0 and f ′(x) > 0 for x > e12, we have that

n1/4 > lnn and therefore1

n5/4>

lnn

n3/2

for sufficiently large n. Since∑ 1

n5/4is a convergent p-series,

∑ lnn

n3/2converges

by the basic comparison test.

56. The series converges if deg q ≥ deg p + 2 and diverges otherwise.

SECTION 12.4

1. converges; ratio test:ak+1

ak=

10k + 1

→ 0 2. converges; root test:(

1k2k

)1/k

=1

2k1/k→ 1

2

3. converges; root test: (ak)1/k =1k→ 0 4. converges; root test: a

1/kk =

k

2k + 1→ 1

2

5. diverges; ratio test:ak+1

ak=

k + 1100

→ ∞ 6. diverges; comparison with∑ 1

k


k8. converges; root test (ak)1/k =

1ln k

→ 0

9. converges; root test: (ak)1/k =23k1/k → 2

310. diverges; comparison with

∑ 1k



SECTION 12.4 643

11. diverges; limit comparison with∑ 1√

k12. converges; limit comparison with

∑ 1k2

13. diverges; ratio test:ak+1

ak=

k + 1104 → ∞ 14. converges; root test: (ak)1/k =

k2/k

e→ 1

e


k3/2

16. converges; ratio test,ak+1

ak=

2k+1(k + 1)!(k + 1)k+1

· kk

2kk!= 2(

k

k + 1

)k

=2(

k+1k

)k → 2e


k2

18. converges; integral test∫ ∞

2

dx

x(lnx)3/2=

2√ln 2

19. diverges; integral test:∫ ∞

2

1x

(lnx)−1/2dx = limb→∞

[2(lnx)1/2

]b2

= ∞


k3/2

21. diverges; divergence test:(

k

k + 100

)k

=(

1 +100k

)−k

→ e−100 �= 0

22. converges; ratio test:[(k + 1)!]2

(2k + 2)!· (2k)!(k!)2

=(k + 1)2

(2k + 1)(2k + 2)→ 1

4


k24. diverges; ak �→ 0


ak=

ln (k + 1)e ln k

→ 1e

26. converges; ratio test:(k + 1)!

(k + 1)k+1· k

k

k!=

1(k+1k

)k → 1e


k3/2

28. converges: ratio test(k + 1)!

1 · 3 · . . . · (2k + 1)· 1 · 3 . . . · (2k − 1)

k!=

k + 12k + 1

→ 12


ak=

2(k + 1)(2k + 1)(2k + 2)

→ 0

30. converges; root test: (ak)1/k =(2k + 1)2

5k2 + 1→ 4

5


ak=

(k + 1)(2k + 1)(2k + 2)(3k + 1)(3k + 2)(3k + 3)

→ 427



644 SECTION 12.4

32. converges by Exercise 38, section 11.2


ak=

1(k + 1)1/2

(k + 1k

)k/2

→ 0 ·√e = 0

34. diverges: ak =(k

9

)k

�→ 0

35. converges; root test: (ak)1/k =√k −

√k − 1 =

1√k +

√k + 1

→ 0

36. converges; root test: (ak)1/k =

k

3k→ 0

37.12

+232

+443

+854

+ · · · =∞∑

k=0

2k

(k + 2)k+1

converges; root test: (ak)1/k =

2(k + 2)1+1/k

→ 0

38. converges: ratio test (see Exercise 28)

39.14

+1 · 34 · 7 +

1 · 3 · 54 · 7 · 10

+ · · · =∞∑

k=0

1 · 3 · · · (1 + 2k)4 · 7 · · · (4 + 3k)

converges; ratio test:ak+1

ak=

3 + 2k7 + 3k

→ 23

40. converges; ratio test :ak+1

ak=

2 · 4 · 6 · . . . · 2(k + 1)3 · 7 · 11 · . . . · [4(k + 1) − 1]

· 3 · 7 · 11 · . . . · (4k − 1)2 · 4 · 6 · . . . · 2k =

2k + 24k + 3

→ 12

41. By the hint∞∑

k=1

k

(110

)k

=110

∞∑k=1

k

(110

)k−1

=110

[1

1 − 1/10

]2=

1081

.

42. (a) If λ > 1, then for k sufficiently large

ak+1

ak> 1 and thus ak+1 > ak

This shows that the kth term cannot tend to 0 and thus the series cannot converge.

(b)∑ 1

kdiverges,

ak+1

ak=

k

k + 1→ 1

∑ 1k2

converges,ak+1

ak=

k2

(k + 1)2→ 1

43. The series∞∑

k=0

k!kk

converges (see Exercise 26). Therefore, limk→∞

k!kk

= 0 by Theorem 12.2.5.



SECTION 12.5 645

44.rn+1

(n + 1)!· n!rn

=r

n + 1→ 0, so by ratio test

∑ rn

n!converges, and therefore

rn

n!→ 0

45. Use the ratio test:

ak+1

ak=

[(k + 1)!]2

[p(k + 1)]!(k!)2

(pk)!

= (k + 1)2(pk)!

(pk)!(pk + 1) · · · (pk + p)=

(k + 1)2

(pk + 1) · · · (pk + p)

Thus

ak+1

ak→

⎧⎨⎩

14, if p = 2

0, if p > 2.

The series converges for all p ≥ 2.

46. By root test: (ak)1/k =r

kr/k=

r(k1/k

)r → r converges if r < 1, diverges if r > 1.

If r = 1, we get∑ 1

k, which diverges.

47. Set bk = akrk. If (ak)1/k → ρ and ρ <

1r, then

(bk)1/k = (akrk)1/k = (ak)1/kr → ρr < 1

and thus, by the root test, Σbk = Σakrk converges.

48. (a)

ak =

⎧⎪⎨⎪⎩

(12)k, k is odd

( 12 )k−2, k is even

⎤⎥⎦

Clearly, (ak)1/k → 12 < 1.

(b) limk→∞

ak+1

akdoes not exist since

ak+1

ak=

⎧⎪⎨⎪⎩

18, k is even

2, k is odd

⎤⎥⎦

SECTION 12.5

1. diverges; ak �→ 0

2. (a)∑

|ak| =∑ 1

2kdiverges, so not absolutely convergent.

(b)1

2(k + 1)<

12k

, ak → 0 : converges conditionally; Theorem 12.5.3.

3. diverges;k

k + 1→ 1 �= 0

4. (a)∑

|ak| =∑ 1

k ln k, does not converge absolutely.

(b) converges conditionally; Theorem 12.5.3.



646 SECTION 12.5

5. (a) does not converge absolutely; integral test,∫ ∞

1

lnx

xdx = lim

b→∞

[12(lnx)2

]b1

= ∞

(b) converges conditionally; Theorem 12.5.3

6. diverges; ak �→ 0


k

another approach:∑(

1k− 1

k!

)=∑ 1

k−∑ 1

k!diverges since

∑ 1k

diverges and∑ 1k!

converges

8. converges absolutely (terms already positive): ratio test,ak+1

ak=

(k + 1)3

2k+1· 2k

k3=(k + 1k

)3

· 12→ 1

2

9. (a) does not converge absolutely; limit comparison with∑ 1

k


10. converges absolutely by ratio test.

11. diverges; ak �→ 0 12. diverges: ak �→ 0

13. (a) does not converge absolutely;

(√k + 1 −

√k) · (

√k + 1 +

√k)

(√k + 1 +

√k)

=1√

k + 1 +√k

and ∑ 1√k +

√k + 1

>∑ 1

2√k + 1

=12

∑ 1√k + 1

(a p-series with p < 1)


14. (a) does not converge absolutely:k

k2 + 1>

k

2k2=

12k

, comparison with∑ 1

2k

(b)k + 1

(k + 1)2 + 1<

k

k2 + 1; converges conditionally; Theorem 12.5.3.

15. converges absolutely (terms already positive); basic comparison,∑sin( π

4k2

)≤∑ π

4k2=

π

4

∑ 1k2

(| sinx| ≤ |x|)

16. (a) does not converges absolutely:∑ 1√

k(k + 1)>∑ 1

k + 1

(b) converges conditionally by Theorem 12.5.3

17. converges absolutely; ratio test,ak+1

ak=

k + 12k

→ 12



SECTION 12.5 647

18. terms all positive, converges absolutely: ak =1√

k√k + 1(

√k +

√k + 1)


k3/2

19. (a) does not converge absolutely; limit comparison with∑ 1

k(b) converges conditionally; Theorem 12.5.3

20. (a) does not converge absolutely:k + 2k2 + k

>k

2k2=

12k


2k(b) converges conditionally; Theorem 12.5.3.

21. diverges; ak =4k−2

ek=

116

(4e

)k

�→ 0

22. converges absolutely by integral test:∫ ∞

1

x22−x dx converges

23. diverges; ak = k sin(1/k) =sin(1/k)

1/k→ 1 �= 0

24. diverges:∣∣∣∣ak+1

ak

∣∣∣∣ = (k + 1)k+1

(k + 1)!· k!kk

=(k + 1k

)k

> 1, so ak �→ 0

25. converges absolutely; ratio test,ak+1

ak=

(k + 1)e−(k+1)

k e−k=

k + 1k

1e→ 1

e

26. (a)∑ cosπk

k=∑ (−1)k

kdoes not converge absolutely.

(b) converges conditionally; Theorem 12.5.3.

27. diverges;∑

(−1)kcosπk

k=∑

(−1)k(−1)k

k=∑ 1

k

28. Converges absolutely; |ak| =∣∣∣∣ sin(πk/2)

k√k

∣∣∣∣ < 1k3/2

29. converges absolutely; basic comparison

∑∣∣∣∣ sin(πk/4)k2

∣∣∣∣ ≤∑ 1k2

30. The series∑(

13k + 2

− 13k + 3

)=∑ 1

(3k + 2)(3k + 3)converges by comparison with

∑ 1k2

.

If∑(

13k + 2

− 13k + 3

− 13k + 4

)converged, then

∑ 13k + 4

=∑(

13k + 2

− 13k + 3

)−∑(

13k + 2

− 13k + 3

− 13k + 4

)would converge, which is

not the case.

31. diverges; ak �→ 0 32. error < a21 =121

∼= 0.04762



648 SECTION 12.5

33. Use (12.5.4); |s− s80| < a81 =1√82

∼= 0.1104 34. error < a5 =1

105= 0.00001

35. Use (12.5.4); |s− s9| < a10 =1

103= 0.001

36. error < an+1 =1

10n+1(a)

110n+1

< 10−3 =⇒ n ≥ 3 (b)1

10n+1< 10−4 =⇒ n ≥ 4

37.1011

; geometric series with a = 1 and r = − 110

, sum =a

1 − r=

1011

38.(0.9)N+1

N + 1< 0.001 N ≥ 32

39. Use (12.5.4); |s− sn| < an+1 =1√n + 2

< 0.005 =⇒ n ≥ 39, 998

40. The series diverges because among the partial sums are all sums of the form12

+13

+14

+ · · · + 1n

Thus for instance,

s1 =12, s5 =

12

+13, s11 =

12

+13

+14, and so on.

This does not violate the theorem on alternating series because, in the notation of the theorem, it is

not true that {ak} decreases.

41. Use (12.5.4).

(a) n = 4;1

(n + 1)!< 0.01 =⇒ 100 < (n + 1)!

(b) n = 6;1

(n + 1)!< 0.001 =⇒ 1000 < (n + 1)!

42. Yes. This can be shown by making slight changes in the proof of Theorem 12.5.3. The even partial

sums s2m are now nonnegative. Since s2m+2 ≤ s2m, the sequence converges; say, s2m → l.

Since s2m+1 = s2m − a2m+1 and a2m+1 → 0, we have s2m+1 → l. Thus, sn → l.

43. No. For instance, set a2k = 2/k and a2k+1 = 1/k.

44. If∑

ak is absolutely convergent, then∑ |ak| converges. Therefore

∑ |bk| by comparison with∑ |ak|. Thus∑

bk is absolutely convergent.

45. (a) Since∑

|ak| converges,∑

|ak|2 =∑

a2k converges (Exercise 49, Section 12.3).

(b)∑ 1

k2converges,

∑(−1)k

1k

is not absolutely convergent.



SECTION 12.6 649

46. s2m+1 = a0 − a1 + a2 − a3 + a4 + · · · − a2m+1

= a0 + (−a1 + a2) + (−a3 + a4) + · · · + (−a2m−1 + a2m) − a2m+1

= a0 + negative terms.

Then s2m+1 < a0 : and {s2m+1} is bounded above.

s2m+3 = s2m+1 + (a2m+2 − a2m+3) > s2m+1, thus {s2m+1} is increasing.

47. See the proof of Theorem 12.8.2.

48. (a)∞∑

k=1

(−1)k−1(a + b) + (a− b)2k

=∞∑

k=1

(−1)k−1(a + b)2k

+∞∑

k=1

a− b

2k

(b) The series is absolutely convergent if a = b = 0; conditionally convergent if a = b �= 0;

divergent if a �= b.

SECTION 12.6

1. −1 + x + 12x

2 − 124x

4 2. 1 + 12x− 1

8x2 + 1

16x3 − 5

128x4

3. − 12x

2 − 112x

4 4. 1 + 12x

2 + 524x

4

5. 1 − x + x2 − x3 + x4 − x5 6. x + x2 + 13x

3 − 130x

5

7. x + 13x

3 + 215x

5 8. x− 12x

5

9. P0(x) = 1, P1(x) = 1 − x, P2(x) = 1 − x + 3x2, P3(x) = 1 − x + 3x2 + 5x3

10. P0(x) = 1, P1(x) = 1 + 3x, P2(x) = 1 + 3x + 3x2, P3(x) = 1 + 3x + 3x2 + x3

11.n∑

k=0

(−1)kxk

k!12.

m∑k=0

x2k+1

(2k + 1)!where m =

n− 12

and n is odd.

13.m∑

k=0

x2k

(2k)!where m =

n

2and n is even 14. −

n∑k=1

xk

k

15. f (k)(x) = rkerx and f (k)(0) = rk, k = 0, 1, 2, . . . . Thus, Pn(x) =n∑

k=0

rk

k!xk

16.m∑

k=0

(−1)k

(2k)!(bx)2k where m =

n

2and n is even.

17. |f(1/2) − P5(1/2)| = |R5(1/2)| ≤ (1)(1/2)6

6!=

1266!

< 0.00002

18. |f(−2) − P7(−2)| = |R7(−2)| ≤ 28

8!< 0.00635



650 SECTION 12.6

19. |f(2) − Pn(2)| = |Rn(2)| ≤ (1)2n+1

(n + 1)!=

2n+1

(n + 1)!; the least integer n that satisfies the inequality

2n+1

(n + 1)!< 0.001 is n = 9.

20. |f(−4) − Pn(−4)| = |Rn(−4)| ≤ (1)4n+1

(n + 1)!=

4n+1

(n + 1)!; the least integer n that satisfies the

inequality4n+1

(n + 1)!< 0.001 is n = 14.

21. |f(1/2) − Pn(1/2)| = |Rn(1/2)| ≤ (3)(1/2)n+1

(n + 1)!=

32n+1(n + 1)!

; the least integer n that satisfies the

inequality3

2n+1(n + 1)!< 0.00005 is n = 9.

22. |f(2) − Pn(2)| = |Rn(2)| ≤ (3)(2)n+1

(n + 1)!=

3 · 2n+1

(n + 1)!; the least integer n that satisfies the inequality

3 · 2n+1

(n + 1)!< 0.0005 is n = 10.

23. |f(x) − P5(x)| = |R5(x)| ≤ (3)|x|66!

=|x|6240

;|x|6240

< 0.05 =⇒ |x|6 < 12 =⇒ |x| < 1.513

24. |f(x) − P9(x)| = |R9(x)| ≤ (3)|x|1010!

=|x|10

1, 209, 600;

|x|101, 209, 600

< 0.05 =⇒ |x|10 < 60480 =⇒|x| < 3.0072

25. The Taylor polynomial

Pn(0.5) = 1 + (0.5) +(0.5)2

2!+ · · · + (0.5)n

n!estimates e0.5 within

|Rn+1(0.5)| ≤ e0.5 |0.5|n+1

(n + 1)!< 2

(0.5)n+1

(n + 1)!.

Since

2(0.5)4

4!=

18(24)

< 0.01,

we can take n = 3 and be sure that

P3(0.5) = 1 + (0.5) +(0.5)2

2+

(0.5)3

6=

7948

differs from√e by less than 0.01. Our calculator gives

7948

∼= 1.645833 and√e ∼= 1.6487213.

26. At x = 0.3 the sine series gives

sin 0.3 = 0.3 − (0.3)3

3!+

(0.3)5

5!− (0.3)7

7!+ · · · .

This is a convergent alternating series with decreasing terms. The first term of magnitude less than

0.01 is (0.3)3/3! = 0.0045. Thus 0.3 differs from sin 0.3 by less than 0.01. Our calculator gives



SECTION 12.6 651

sin 0.3 ∼= 0.2955202. The estimate

0.3 − (0.3)3

3!= 0.2955

is much more accurate. The series converges very rapidly for small values of x.

27. At x = 1, the sine series gives

sin 1 = 1 − 13!

+15!

− 17!

+ · · · .


0.01 is 1/5! = 1/110. Thus

1 − 13!

= 1 − 16

=56

differs from sin 1 by less than 0.01. Our calculator gives

56∼= 0.8333333 and sin 1 ∼= 0.84114709.

The estimate

1 − 13!

+15!

=101110

∼= 0.8416666

is much more accurate.

28. At x = 1.2 the logarithm series (12.6.8) gives

ln 1.2 = ln(1 + 0.2) = 0.2 − 12(0.2)2 +

13(0.2)3 − 1

4(0.2)4 + · · · .


0.01 is (0.2)3/3 ∼= 0.00267. Thus

0.2 − 12(0.2)2 = 0.18

differs from ln 1.2 by less than 0.01. Our calculator gives ln 1.2 ∼= 0.1823215.

29. At x = 1, the cosine series gives

cos 1 = 1 − 12!

+14!

− 16!

+18!

+ · · · .


0.01 is 1/6! = 1/720. Thus

1 − 12!

+14!

= 1 − 12

+124

=1324

differs from cos 1 by less than 0.01. Our calculator gives

1324

∼= 0.5416666 and cos 1 ∼= 0.5403023.

30. The Taylor polynomial

Pn(0.8) = 1 + (0.8) +(0.8)2

2!+ · · · + (0.8)n

n!



652 SECTION 12.6

estimates e0.8 within

|Rn(0.8)| ≤ e0.8 (0.8)n+1

(n + 1)!< 3

(0.8)n+1

(n + 1)!.

Since 3(0.8)5

5!< 0.0082 < 0.01 we can take n = 4 and be sure that

P4(0.8) = 1 + (0.8) +(0.8)2

2!+

(0.8)3

3!+

(0.8)4

4!= 2.224

differs from e0.8 by less than 0.01. Our calculator gives e0.8 ∼= 2.2255409.

31. First convert 10◦ to radians: 10◦ =10180

π ∼= 0.1745 radians

At x = 0.1745, the sine series gives

sin 0.1745 = 0.1745 − (0.1745)3

3!+

(0.1745)5

5!− · · · .


0.01 is (0.1745)3/3! ∼= 0.00089. Thus 0.1745 differs from sin 10◦ by less than 0.01. Our calculator gives

sin 10◦ ∼= 0.1736

32. At x = 6◦ =π

30, the cosine series gives cos

π

30= 1 − 1

2

( π

30

)2

+14!

( π

30

)4

− 16!

( π

30

)6

+ · · ·The first term less than 0.01 is 1

2

(π30

)2 ∼= 0.0055, so 1 differs from cos 6◦ by less than 0.01.

Calculator gives cos 6◦ ∼= 0.9945219

33. f(x) = e2x; f (5)(x) = 25e2x; R4(x) =25e2c

5!x5 =

415

e2cx5, where c is between 0 and x.

34. Rn(x) = R5(x) =f6(c)

(5 + 1)!x6 =

−120(1 + c)−6

6!x6 = −1

6

(x

1 + c

)6

, where c is between 0 and x.

35. f(x) = cos 2x; f (5)(x) = −25 sin 2x

R4(x) =−25 sin 2c

5!x5 = − 4

15sin(2c)x5,

where c is between 0 and x.

36. Rn(x) = R3(x) =f4(c)

4!x4 =

− 1516 (c + 1)−7/2

4!x4 =

−5x4

128(c + 1)7/2, where c is between 0 and x.

37. f(x) = tanx; f ′′′(x) = 6 sec4 x− 4 sec2 x

R2(x) =6 sec4 c− 4 sec2 c

3!x3 =

3 sec4 c− 2 sec2 c

3x3,


38. Rn(x) = R5(x) =f6(c)

6!x6 =

− sin c

6!x6, where c is between 0 and x.



SECTION 12.6 653

39. f(x) = arctan x; f ′′′(x) =6x2 − 2

(1 + x2)3

R2(x) =6c2 − 2

3! (1 + c2)3x3 =

3c2 − 13 (1 + c2)3

x3,


40. Rn(x) = R4(x) =f5(c)

5!x5 =

−120(1 + c)−6

5!x5 =

−x5

(1 + c)6, where c is between 0 and x.

41. f(x) = e−x; f (k)(x) = (−1)ke−x, k = 0, 1, 2, . . .

Rn(x) =(−1)n+1e−c

(n + 1)!xn+1,


42. Rn(x) =fn+1(c)(n + 1)!

xn+1 =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

(−1)n−12 2n+1 cos 2c(n + 1)!

xn+1 n odd

(−1)n2 2n+1 sin 2c(n + 1)!

xn+1 n even, where c is between 0 and x.

43. f(x) =1

1 − x; f (k)(x) =

k!(1 − x)k+1

, k = 0, 1, 2, . . .

Rn(x) =(n + 1)!

(1 − c)n+2(n + 1)!xn+1 =

1(1 − c)n+2

xn+1, where c is between 0 and x.

44. Rn(x) =fn+1(c)(n + 1)!

xn+1 =(−1)n+1(n!)/(1 + c)n+1

(n + 1)!xn+1 =

(−1)n+1

n + 1

(x

1 + c

)n+1

,


45. By (12.6.8)

Pn(x) = x− x2

2+

x3

3− x4

4+ · · · + (−1)n+1x

n

n.

For 0 ≤ x ≤ 1 we know from (12.5.4) that

|Pn(x) − ln (1 + x)| < xn+1

n + 1.

(a) n = 4;(0.5)n+1

n + 1≤ 0.01 =⇒ 100 ≤ (n + 1)2n+1 =⇒ n ≥ 4

(b) n = 2;(0.3)n+1

n + 1≤ 0.01 =⇒ 100 ≤ (n + 1)

(103

)n+1

=⇒ n ≥ 2

(c) n = 999;(1)n+1

n + 1≤ 0.001 =⇒ 1000 ≤ n + 1 =⇒ n ≥ 999

46. (a) Since17

7!∼= 0.0002, use P5.

(b) Since211

11!∼= 0.00005 is the first term less than 0.001, use P9.

(c) Since313

13!∼= 0.0002 is the first term less than 0.001, use P11.



654 SECTION 12.6

47. f(x) = ex; f (n)(x) = ex; Rn(x) =ec

(n + 1)!xn+1, |c| < |x|

(a) We want |Rn(1/2)| < .00005 : for 0 < c < 11 , we have

|Rn(1/2)| =ec

(n + 1)!

(12

)n+1

<e1/2

(n + 1)!

(12

)n+1

<2

2n+1(n + 1)!< 0.00005

You can verify that this inequality is satisfied if n ≥ 5.

P5(x) = 1 + x +x2

2!+

x3

3!+

x4

4!+

x5

5!

P5(1/2) = 1 +12

+18

+148

+1

384+

13840

∼= 1.6487

(b) We want |Rn(−1)| < .0005 : for −1 < c < 0, we have

|Rn(−1)| =ec

(n + 1)!

∣∣(−1)n+1∣∣ < 1

(n + 1)!< 0.0005

You can verify that this inequality is satisfied if n ≥ 7.

P7(x) =7∑

k=0

xk

k!; P7(−1) =

7∑k=0

(−1)k

k!∼= 0.368

48. (a)14!

( π

30

)4

is the first term less than 0.0005, so use P2

(π30

)= 1 − 1

2

(π30

)2 ∼= 0.995

(b) 9◦ =π

2014!

( π

20

)4

is the first term less than 0.0005, so use P2

(π20

)= 1 − 1

2

(π20

)2 ∼= 0.9877

49. The result follows from the fact that P (k)(0) =

{k!ak, 0 ≤ k ≤ n

0, n < k

].

50. Straightforward

51.dk

dxk(sinhx) =

{sinhx, if k is odd

coshx, if k is even

Thusdk

dxk(sinhx)

∣∣x=0

=

{0, if k is odd

1, if k is evenand

sinhx = x +x3

3!+

x5

5!+ · · · =

∞∑k=0

x(2k+1)

(2k + 1)!

52.coshx =

12(ex + e−x) =

12

∞∑n=0

xn

n!+

12

∞∑n=0

(−x)n

n!

=∞∑

k=0

x2k

(2k)!because the odd terms cancel out

53. Set t = ax. Then, eax = et =∞∑

k=0

tk

k!=

∞∑k=0

ak

k!xk, (−∞,∞).



SECTION 12.6 655

54. sin ax =∞∑

k=0

(−1)k

(2k + 1)!(ax)2k+1 =

∞∑k=0

(−1)ka2k+1

(2k + 1)!x2k+1; (−∞,∞)

55. Set t = ax. Then, cos ax = cos t =∞∑

k=0

(−1)k

(2k)!t2k =

∞∑k=0

(−1)ka2k

(2k)!x2k, (−∞,∞).

56. ln(1 − ax) =∞∑

k=1

(−1)k+1

k(−ax)k = −

∞∑k=1

ak

kxk;

[−1a,1a

)

57. See (12.5.8): ln(a + x) = ln[a(1 +

x

a

)]= ln a + ln

(1 +

x

a

)= ln a +

∞∑k=1

(−1)k+1

kakxk.

By (12.5.8) the series converges for −1 <x

a≤ 1; that is, −a < x ≤ a.

58. f(x) = ln(

1 + x

1 − x

)= ln(1 + x) − ln(1 − x); f(0) = 0

f ′(x) =1

1 + x+

11 − x

, f ′(0) = 2

f ′′(x) =−1

(1 + x)2+

1(1 − x)2

, f ′′(0) = 0

f ′′′(x) =2

(1 + x)3+

2(1 − x)3

, f ′′′(0) = 4

In general, fn(x) =(−1)n+1(n− 1)!

(1 + x)n+

(n− 1)!(1 − x)n

, f (n)(0) = 2(n− 1)! for n odd,

0 for n even The result follows.

59. ln 2 = ln(

1 + 1/31 − 1/3

)∼= 2

[13

+13

(13

)3

+15

(13

)5]

=8421115

.

Our calculator gives8421115

∼= 0.6930041 and ln 2 ∼= 0.6931471.

60.1 + x

1 − x= 1.4 gives x =

16; ln 1.4 ∼= 2

[16

+13

(16

)3]

=109324

∼= 0.336

61. Set u = (x− t)k, dv = f (k+1)(t) dt

du = −k(x− t)k−1 dt, v = f (k)(t).

Then, − 1k!

∫ x

0

f (k+1)(t)(x− t)k dt

= − 1k!

[(x− t)kf (k)(t)

]x0− 1

k!

∫ x

0

k(x− t)k−1f (k)(t) dt

=f (k)(0)

k!xk − 1

(k − 1)!

∫ x

0

f (k)(t)(x− t)k−1 dt.

The given identity follows.



656 SECTION 12.6

62. Suppose x > 0. By the second mean-value theorem for integrals (Theorem 5.9.3), there is at least one

number c ∈ (0, x) such that

1n!

∫ x

0

f (n+1)(t)(x− t)ndt =f (n+1)(c)

n!

∫ x

0

(x− t)n dt

=fn+1(c)

n!

[− (x− t)n+1

n + 1

]x0

=fn+1(c)(n + 1)!

xn+1.

The same result follows if x < 0; see Section 5.9, Exercise 32.

If x > 0, then

|Rn(x)| =1n!

∣∣∣∣∫ x

0

f (n+1)(t)(x− t)n dt∣∣∣∣ ≤ 1

n!

∫ x

0

∣∣∣f (n+1)(t)∣∣∣ (x− t)n dt

≤ 1n!

∫ x

0

M(x− t)n dt where M = maxt∈I

|fn+1(t)|

=M

n!

∫ x

0

(x− t)n dt =M

n!

[−(x− t)n+1

n + 1

]x0

= M|x|n+1

(n + 1)!.

If x < 0, then

|Rn(x)| =1n!

∣∣∣∣∫ x

0

f (n+1)(t)(x− t)n dt∣∣∣∣ ≤ 1

n!

∫ 0

x

∣∣∣f (n+1)(t)∣∣∣ (t− x)n dt

=M

n!

[(t− x)n+1

n + 1

]0x

= M|x|n+1

(n + 1)!.

63. (a)

−5 5x

0.5

1

y

(b) Let g(x) =x−n

e1/x2 . Then limx→0

g(x) has the form ∞/∞. Successive applications of L’Hopital’s

rule will finally produce a quotient of the formcxk

e1/x2 , where k is a nonnegative integer and c

is a constant. It follows that limx→0

g(x) = 0.

(c) f ′(0) = limx→0

e−1/x2 − 0x

= 0 by part (b). Assume that f (k)(0) = 0. Then

f (k+1)(0) = limx→0

f (k)(x) − 0x

= limx→0

f (k)(x)x

.

Now, f (k)(x)/x is a sum of terms of the form ce−1/x2/xn, n a positive integer and c a constant.

Again by part (b), f (k+1)(0) = 0. Therefore, f (n)(0) = 0 for all n.

(d) 0 (e) x = 0



SECTION 12.7 657

64.

x

y

P2

P4

cos x

x

y

P6

P8

cos x

65.

−1 1 2x

−1

1

y

P2

P3

f

P4

P5

−1 1x

−1

1

y

P5 P3

f

P4

P2

66. (1) 0 < q!(e− sq) = (q!)∞∑

k=q+1

1k!

=q!

(q + 1)!+

q!(q + 2)!

+q!

(q + 3)!+ · · ·

≤ 1q + 1

+1

(q + 1)2+

1(q + 3)3

+ · · ·

≤ 1q + 1

[1 +

1q + 1

+1

(q + 1)2+ · · ·

](geometric series)

=1

q + 1

[1

1 − 1/(q + 1)

]=

1q

(2) If e equaled p/q, then q!e would be an integer, and since q!sq is an integer, so would q!(e− sq).

But 0 < q!(e− sq) <1q< 1, impossible.

SECTION 12.7

1. f(x) =√x = x1/2; f(4) = 2

f ′(x) =12x−1/2; f ′(4) =

14

f ′′(x) = − 14x−3/2; f ′′(4) = − 1

32



658 SECTION 12.7

f ′′′(x) =38x−5/2; f ′′′(4) =

3256

f (4)(x) = − 1516

x−7/2

P3(x) = 2 +14(x− 4) − 1/32

2!(x− 4)2 +

3/2563!

(x− 4)3

= 2 +14(x− 4) − 1

64(x− 4)2 +

1511

(x− 4)3

R3(x) =f (4)(c)

4!(x− 4)4 = −15

16· 14!c−7/2(x− 4)4 = − 5

118c7/2(x− 4)4, where c is between 4 and x.

2. f(π

3

)=

12, f ′

(π3

)= − sin

π

3= −

√3

2, f ′′

(π3

)= − cos

π

3= −1

2,

f ′′′(π

3

)= sin

π

3=

√3

2, f (4)

(π3

)= cos

π

3=

12, f (5) (c) = − sin c

P4(x) =12−

√3

2

(x− π

3

)− 1

4

(x− π

3

)2

+√

32 · 3!

(x− π

3

)3

+1

2 · 4!

(x− π

3

)4

R4(x) =− sin c

5!

(x− π

3

)5

, where c is between π/3 and x.

3. f(x) = sinx; f(π/4) =√

22

f ′(x) = cosx; f ′(π/4) =√

22

f ′′(x) = − sinx; f ′′(π/4) = −√

22

f ′′′(x) = − cosx; f ′′′(π/4) = −√

22

f (4)(x) = sinx; f (4)(π/4) =√

22

f (5)(x) = cosx

P4(x) =√

22

+√

22

(x− π

4

)−

√2/22!

(x− π

4

)2

−√

2/23!

(x− π

4

)3

+√

2/24!

(x− π

4

)4

=√

22

+√

22

(x− π

4

)−

√2

4

(x− π

4

)2

−√

211

(x− π

4

)3

+√

248

(x− π

4

)4

R4(x) =f (5)(c)

5!

(x− π

4

)5

=cos c110

(x− π

4

)5

, where c is between π/4 and x.

4. f (1) = 0, f ′ (1) =11

= 1, f ′′ (1) =−112

= −1,

f ′′′ (1) =213

= 2, f (4) (1) =−614

= −6, f (5) (1) =4!15

= 4!, f (6)(c) =−5!c6

P5(x) = (x− 1) − 12(x− 1)2 +

13(x− 1)3 − 1

4(x− 1)4 +

15(x− 1)5

R5(x) =−5!c6

· 16!

(x− 1)6 = −16

(x− 1c

)6

, where c is between 1 and x.



SECTION 12.7 659

5. f(x) = arctan (x) f(1) =π

4

f ′(x) =1

1 + x2f ′(1) = 1

2

f ′′(x) =−2x

(1 + x2)2f ′′(1) = − 1

2

f ′′′(x) =6x2 − 2

(1 + x2)3f ′′′(1) = 1

2

f (4)(x) =24(x− x3)(1 + x2)4

P3(x) =π

4+

12(x− 1) − 1/2

2!(x− 1)2 +

1/23!

(x− 1)3 =π

4+

12(x− 1) − 1

4(x− 1)2 +

112

(x− 1)3

R3(x) =f (4)(c)

4!(x− 1)4 =

24(c− c3)(1 + c2)4

· 14!

(x− 1)4 =c− c3

(1 + c2)4(x− 1)4, where c is between 1 and x.

6. f

(12

)= 0, f ′

(12

)= −π sin

π

2= −π, f ′′

(12

)= −π2 cos

π

2= 0,

f ′′′(

12

)= π3 sin

π

2= π3, f (4)

(12

)= π4 cos

π

2= 0, f (5) (c) = −π5 sinπc

P4(x) = −π

(x− 1

2

)+

π3

3!

(x− 1

2

)3

; R4(x) =−π5 sinπc

5!

(x− 1

2

)5

,

where c is between 1/2 and x.

7. g(x) = 6 + 9(x− 1) + 7(x− 1)2 + 3(x− 1)3, (−∞,∞)

8. 11 + 23(x− 2) + 19(x− 2)2 + 7(x− 2)3 + (x− 2)4; (−∞,∞)

9. g(x) = −3 + 5(x + 1) − 19(x + 1)2 + 20(x + 1)3 − 10(x + 1)4 + 2(x + 1)5, (−∞,∞)

10.1x

=1

1 + (x− 1)=

∞∑k=0

(−1)k(x− 1)k; (0, 2)

11. g(x) =1

1 + x=

12 + (x− 1)

=12

⎡⎢⎢⎣ 1

1 +(x− 1

2

)⎤⎥⎥⎦ =

12

∞∑k=0

(−1)k(x− 1

2

)k

(geometric series)

=∞∑

k=0

(−1)k(x− 1)k

2k+1for

∣∣∣∣x− 12

∣∣∣∣ < 1 and thus for − 1 < x < 3

12.1

b + x=

1b + a + x− a

=1

b + a· 11 + x−a

b+a

=1

b + a

∞∑k=0

(−1)k(x− a

b + a

)k

=∞∑

k=0

(−1)k(

1a + b

)k+1

(x− a)k, (a− |a + b|, a + |a + b|)



660 SECTION 12.7

13. g(x) =1

1 − 2x=

15 − 2(x + 2)

=15

[1

1 − 25 (x + 2)

]=

15

∞∑k=0

[25(x + 2)

]k

=∞∑

k=0

2k

5k+1(x + 2)k for

∣∣∣∣25(x + 2)∣∣∣∣ < 1 and thus for − 9

2< x <

12

14. e−4x = e−4(x+1)e4 = e4∞∑

k=0

(−1)k4k

k!(x + 1)k; (−∞,∞)

15. g(x) = sinx = sin [(x− π) + π] = sin (x− π) cosπ + cos (x− π) sinπ

= − sin (x− π) = −∞∑

k=0

(−1)k(x− π)2k+1

(2k + 1)!

(12.6.8)∧

=∞∑

k=0

(−1)k+1 (x− π)2k+1

(2k + 1)!, (−∞,∞)

16. sinx = cos(x− π

2

)=

∞∑k=0

(−1)k

(2k)!

(x− π

2

)2k

; (−∞,∞)

17. g(x) = cosx = cos [(x− π) + π] = cos (x− π) cosπ − sin (x− π) sinπ

= − cos (x− π) = −∞∑

k=0

(−1)k(x− π)2k

(2k)!=

∞∑k=0

(−1)k+1 (x− π)2k

(2k)!, (−∞,∞)

(12.6.7)∧

18. cosx = − sin(x− π

2

)=

∞∑k=0

(−1)k+1

(2k + 1)!

(x− π

2

)2k+1

; (−∞,∞)

19. g(x) = sin12πx = sin

[π2

(x− 1) +π

2

]= sin

[π2

(x− 1)]cos

π

2+ cos

[π2

(x− 1)]sin

π

2

= cos[π2

(x− 1)]

=∞∑

k=0

(−1)k(π

2

)2k (x− 1)2k

(2k)!, (−∞,∞)

(12.6.7)∧

20. sinπx = − sinπ(x− 1) = −∞∑

k=0

(−1)k

(2k + 1)![π(x− 1)]2k+1 =

∞∑k=0

(−1)k+1π2k+1

(2k + 1)!(x− 1)2k+1; (−∞,∞)



SECTION 12.7 661

21. g(x) = ln (1 + 2x) = ln [3 + 2(x− 1)] = ln[3(1 + 2

3 (x− 1))]

= ln 3 + ln[1 +

23(x− 1)

]= ln 3 +

∞∑k=1

(−1)k+1

k

[23(x− 1)

]k

(12.6.8)∧

= ln 3 +∞∑

k=1

(−1)k+1

k

(23

)k

(x− 1)k.

This result holds if −1 < 23 (x− 1) ≤ 1, which is to say, if − 1

2 < x ≤ 52 .

22. ln(2 + 3x) = ln [14 + 3(x− 4)] = ln 14 + ln[1 +

314

(x− 4)]

= ln 14 +∞∑

k=1

(−1)k+1

k

(314

)k

(x− 4)k;

(−2

3,263

]

23. g(x) = x ln x

g′(x) = 1 + ln x

g′′(x) = x−1

g′′′(x) = −x−2

g(iv)(x) = 2x−3

...

g(k)(x) = (−1)k(k − 2)!x1−k, k ≥ 2.

Then, g(2) = 2 ln 2, g′(2) = 1 + ln 2, and g(k)(2) =(−1)k(k − 2)!

2k−1, k ≥ 2.

Thus, g(x) = 2 ln 2 + (1 + ln 2)(x− 2) +∞∑

k=2

(−1)k

k(k − 1)2k−1(x− 2)k.

24. g(2) = 4 + e6; g′(x) = 2x + 3e3x, g′(2) = 4 + 3e6, g′′(x) = 2 + 9e3x,

g′′(2) = 2 + 9e6, g′′′(x) = 27e3x, g′′′(2) = 27e6, gn(x) = 3ne3x

=⇒ g(x) = (4 + e6) + (4 + 3e6)(x− 2) + (1 +92e6)(x− 2)2 + e6

∞∑k=3

3k

k!(x− 2)k

25. g(x) = x sinx = x

∞∑k=0

(−1)kx2k+1

(2k + 1)!=

∞∑k=0

(−1)kx2k+2

(2k + 1)!

26. ln(x2) = 2 lnx = 2 ln[1 + (x− 1)] = 2∞∑

k=1

(−1)k+1

k(x− 1)k



662 SECTION 12.7

27. g(x) = (1 − 2x)−3

g′(x) = −2(−3)(1 − 2x)−4

g′′(x) = (−2)2(4 · 3)(1 − 2x)−5

g′′′(x) = (−2)3(−5 · 4 · 3)(1 − 2x)−6

...

g(k)(x) = (−2)k[(−1)k

(k + 2)!2

](1 − 2x)−k−3, k ≥ 0.

Thus, g(k)(−2) = (−2)k[(−1)k

(k + 2)!2

]5−k−3 =

2k−1

5k+3(k + 2)!

and g(x) =∞∑

k=0

(k + 2)(k + 1)2k−1

5k+3(x + 2)k.

28. sin2 x =12− cos 2x

2=

12

+12

cos 2(x− π

2) =

12

+12

∞∑k=0

(−1)k

(2k)!22k(x− π

2)2k

= 1 +∞∑

k=1

(−1)k22k−1

(2k)!

(x− π

2

)2k

29. g(x) = cos2 x =1 + cos 2x

2=

12

+12

cos [2(x− π) + 2π]

=12

+12

cos [2(x− π)] =12

+12

∞∑k=0

(−1)k[2(x− π)]2k

(2k)!

= 1 +∞∑

k=1

(−1)k22k−1

(2k)!(x− π)2k

(k = 0 term is 12 )

30. g(x) = (1 + 2x)−4, g′(x) = −4(1 + 2x)−5 · 2, g′′(x) = 20(1 + 2x)−6 · 4, g′′′(x) = −110(1 + 2x)−7 · 23,

gn(x) = (−1)n(n + 3)!

3!· 2n · (1 + 2x)−n−4 gn(2) = (−1)n

(n + 3)!3!

· 2n

5n+4.

g(x) =∞∑

k=0

(−1)k

k!· (k + 3)!

3!· 2k

5k+4· (x− 2)k =

∞∑k=0

(−1)k

3(k + 3)(k + 2)(k + 1)

2k−1

5k+4(x− 2)k

31. g(x) = xn

g′(x) = nxn−1

g′′(x) = n(n− 1)xn−2

g′′′(x) = n(n− 1)(n− 2)xn−3

...

g(k)(x) = n(n− 1) · · · (n− k + 1)xn−k, 0 ≤ k ≤ n

g(k)(x) = 0, k > n.



SECTION 12.7 663

Thus,

g(k)(1) =

⎧⎨⎩

n!(n− k)!

, 0 ≤ k ≤ n

0, k > n

⎤⎦ and g(x) =

n∑k=0

n!(n− k)!k!

(x− 1)k.

32. (x− 1)n =n∑

k=0

n!(n− k)!k!

xk(−1)n−k

33. (a)ex

ea= ex−a =

∞∑k=0

(x− a)k

k!, ex = ea

∞∑k=0

(x− a)k

k!

(b) ea+(x−a) = ex = ea∞∑

k=0

(x− a)k

k!, ex1+x2 = ex1

∞∑k=0

xk2

k!= ex1ex2

(c) e−a∞∑

k=0

(−1)k(x− a)k

k!

34. (a) sinx = sin a + (x− a) cos a− (x− a)2

2!sin a +

(x− a)3

3!cos a + · · ·

cosx = cos a− (x− a) sin a− (x− a)2

2!cos a +

(x− a)3

3!sin a + · · ·

(b) in both instances∞∑

k=0

|ak| ≤∞∑

k=0

|x− a|kk!

(c)

sin(x1 + x2)

= sinx1 + x2 cosx1 −(x2)2

2!sinx1 −

(x2)3

3!cosx1 + · · ·

=(

sinx1 −x2

2

2!sinx1 +

x24

4!sinx1 − · · ·

)+(x2 cosx1 −

x22

3!cosx1 +

x25

5!cosx1 − · · ·

)

= sinx1

(1 − x2

2

2!+

x24

4!− · · ·

)+ cosx1

(x2 −

x2

3!+

x25

5!− · · ·

)= sinx1 cosx2 + cosx1 sinx2

The other formula can be derived in a similar manner.

35. P6(x) =π

4+

12(x− 1) − 1

4(x− 1)2 +

112

(x− 1)3 − 140

(x− 1)5 +148

(x− 1)6

36. P8(x) = cosh 4 + 2 sinh 4(x− 2) + 2 cosh 4(x− 2)2 +43

sinh 4(x− 2)3 + · · · + 2315

cosh 4(x− 2)8



664 SECTION 12.8

SECTION 12.8

1. (a) converges (b) absolutely converges (c) ? (d) diverges

2. (a) absolutely converges (b) diverges (c) ? (d) convergent

3. (−1, 1); ratio test:bk+1

bk=

k + 1k

|x| → |x|, series converges for |x| < 1.

At the endpoints x = 1 and x = −1 the series diverges since at those points bk �→ 0.

4. [−1, 1); ratio test:∣∣∣∣ xk+1

k + 1· k

xk

∣∣∣∣ = k

k + 1|x| → |x| =⇒ r = 1

At x = 1 :∑ 1

k, diverges; at x = −1,

∑ (−1)k

kconverges.

5. (−∞,∞); ratio test:bk+1

bk=

|x|(2k + 1)(2k + 2)

→ 0, series converges all x.

6.[−1

2,12

]; root test:

∣∣∣∣2kk2xk

∣∣∣∣1/k =2 · |x|k2/k

→ 2|x| =⇒ r =12

At x =12

:∑ 1

k2, converges; at x = −1

2:∑ (−1)k

k2converges.

7. Converges only at 0; divergence test: (−k)2kx2k → 0 only if x = 0, and series

clearly converges at x = 0.

8. (−1, 1]; root test:∣∣∣∣ xk

√k

∣∣∣∣1/k =|x|kk/2

→ |x| =⇒ r = 1

At x = 1 :∑ (−1)k√

k, converges; at x = −1 :

∑ 1√k

diverges.

9. [−2, 2); root test: (bk)1/k =|x|

2k1/k→ |x|

2, series converges for |x| < 2.

At x = 2 series becomes∑ 1

k, the divergent harmonic series.

At x = −2 series becomes∑

(−1)k1k, a convergent alternating series.

10. [−2, 2]; root test:∣∣∣∣ xk

k22k

∣∣∣∣1/k =|x|

k2/k2→ |x|

2=⇒ r = 2

At x = 2 :∑ 1

k2, converges; at x = −2 :

∑ (−1)k

k2converges.

11. Converges only at 0; divergence test:(

k

100

)k

xk → 0 only if x = 0, and series

clearly converges at x = 0.



SECTION 12.8 665

12. (−1, 1); ratio test:(k + 1)2|x|k+1

1 + (k + 1)2· 1 + k2

k2|x|k → |x| =⇒ r = 1

At x = 1 :∑ k2

1 + k2, diverges; at x = −1 :

∑ (−1)kk2

1 + k2diverges.

13.[−1

2,12

); root test: (bk)1/k =

2|x|√k1/k

→ 2|x|, series converges for |x| < 12.

At x =12

series becomes∑ 1√

k, a divergent p-series.

At x = −12

series becomes∑

(−1)k1√k, a convergent alternating series.

14. [−1, 1); ratio test:|x|k+1

ln(k + 1)· ln k

|x|k = |x| ln k

ln(k + 1)→ |x| =⇒ r = 1

At x = 1 :∑ 1

ln k, diverges; at x = −1 :

∑ (−1)k

ln kconverges.

15. (−1, 1); ratio test:bk+1

bk=

k2

(k + 1)(k − 1)|x| → |x|, series converges for |x| < 1.

At the endpoints x = 1 and x = −1 the series diverges since there bk �→ 0.

16.(− 1|a| ,

1|a|

); root test:

∣∣∣∣kakxk

∣∣∣∣1/k = k1/k|a| · |x| → |a| · |x| =⇒ r =1|a|

At x =1|a| :

∑kak

1|a|k , diverges, similarly at x = − 1

|a| .

17. (−10, 10); root test: (bk)1/k =k1/k

10|x| → |x|

10, series converges for |x| < 10.


18. (−e, e); root test:∣∣∣∣3k2xk

ek

∣∣∣∣1/k =31/kk2/k

e|x| → |x|

e=⇒ r = e

Diverges at x = ±e.

19. (−∞,∞); root test: (bk)1/k =|x|k

→ 0, series converges for all x.

20. (−∞,∞); ratio test:7k+1|x|k+1

(k + 1)!· k!7k|x|k =

7|x|k + 1

→ 0, =⇒ r = ∞

21. (−∞,∞); root test: (bk)1/k =|x− 2|

k→ 0, series converges all x.

22. converges only at 0, ratio test:(k + 1)!|x|k+1

k!|x|k = (k + 1)|x| → ∞ =⇒ r = 0



666 SECTION 12.8

23.(−3

2,32

); ratio test:

bk+1

bk=

2k+1

3k+2|x|

2k

3k+1

=23|x|, series converges for |x| < 3

2.

At the endpoints x = 3/2 and x = −3/2, the series diverges since there bk �→ 0.

24. (−∞,∞); ratio test:2k+1|x|k+1

(2k + 2)!· (2k)!2k|x|k =

2|x|(2k + 1)(2k + 2)

→ 0 =⇒ r = ∞

25. Converges only at x = 1; ratio test:bk+1

bk=

k3

(k + 1)2|x− 1| → ∞ if x �= 1

The series clearly converges at x = 1; otherwise it diverges.

26.[−1e,1e

]root test:

∣∣∣∣ (−e)k

k2xk

∣∣∣∣1/k =e|x|k2/k

→ e|x| =⇒ r =1e.

Converges at x = ±1e;

27. (−4, 0); ratio test:bk+1

bk=

k2 − 12k2

|x + 2| → |x + 2|2

, series converges for |x + 2| < 2.

At the endpoints x = 0 and x = −4, the series diverges since there bk �→ 0.

28. [−2, 0); ratio test:ln(k + 1)k + 1

|x + 1|k+1 · k

ln k|x + 1|k =ln(k + 1)

ln k· k

k + 1|x + 1| → |x + 1| =⇒ r = 1

At x = 0,∑ ln k

kdiverges, at x = −2,

∑ ln k

k(−1)k converges.


bk=

(k + 1)2

k2(k + 2)|x + 3| → 0, series converges for all x.

30. (4 − e, 4 + e); root test:∣∣∣∣k3

ek(x− 4)k

∣∣∣∣1/k =k3/k

e|x− 4| → |x− 4|

e=⇒ r = e

At x = 4 + e,∑

k3 diverges, at x = 4 − e,∑

(−1)kk3 diverges.

31. (−1, 1); root test: (bk)1/k =(

1 +1k

)|x| → |x|, series converges for |x| < 1.

At the endpoints x = 1 and x = −1, the series diverges since there bk �→ 0

[ recall(

1 +1k

)k

→ e]

32.[a− 1

|a| , a +1|a|

]; root test:

∣∣∣∣ (−1)kak

k2(x− a)k

∣∣∣∣1/k =|a|k2/k

|x− a| → |a| · |x− a| =⇒ r =1|a|

Converges at both a− 1|a| , a +

1|a|

(compare with

∑ 1k2

).



SECTION 12.8 667

33. (0, 4); ratio test:bk+1

bk=

ln (k + 1)ln k

|x− 2|2

→ |x− 2|2

, series converges for |x− 2| < 2.

At the endpoints x = 0 and x = 4 the series diverges since there bk �→ 0.

34. (−∞,∞); root test:|x− 1|ln k

→ 0 =⇒ r = ∞

35.(−5

2,12

); root test: (bk)1/k = 2

3 |x + 1| → 23 |x + 1|, series converges for |x + 1| < 3

2 .

At the endpoints x = − 52 and x = 1

2 the series diverges since there bk �→ 0.

36.[2 − 1

π, 2 +

1π

]; ratio test:

21

k+1πk+1|x− 2|k+1

(k + 1)(k + 2)(k + 3)· k(k + 1)(k + 2)

21k πk|x− 2|k

→ π|x− 2| =⇒ r =1π

Converges at x = 2 ± 1π

(compare with

∑ 1k3

).

37. 1 − x

2+

2x2

4− 3x3

8+

4x4

16− · · · = 1 +

∞∑k=1

(−1)kkxk

2k

(−2, 2); ratio test:bk+1

bk=

k + 12k

|x| → |x|2, series converges for |x| < 2.


38.152

(x− 1) +454

(x− 1)2 +956

(x− 1)3 + · · · =∞∑

k=1

k2

52k(x− 1)k.

(−24, 26); root test:∣∣∣∣ k2

52k(x− 1)k

∣∣∣∣1/k =k2/k

52|x− 1| → |x− 1|

25=⇒ r = 25

At x = −24,∑

(−1)kk2 diverges, at x = 26,∑

k2 diverges.

39.3x2

4+

9x4

9+

27x6

16+

81x8

25+ · · · =

∞∑k=1

3k

(k + 1)2x2k

[− 1√

3,

1√3

]; ratio test:

bk+1

bk=

3(k + 1)2

(k + 2)2x2 → 3x2, series converges for x2 <

13

or |x| < 1√3.

At x = ± 1√3, the series becomes

∑ 1(k + 1)2

∼=∑ 1

n2, a convergent series p-series.

40.116

(x + 1) − 225

(x + 1)2 +336

(x + 1)3 + · · · =∞∑

k=1

(−1)k+1k

(k + 3)2(x + 1)k.

(−2, 0]; ratio test:(k + 1)|x + 1|k+1

(k + 4)2· (k + 3)2

k|x + 1|k → |x + 1| =⇒ r = 1

At x = 0,∑ (−1)k+1k

(k + 3)2converges; at x = −2, −

∑ k

(k + 3)2diverges.



668 SECTION 12.8

41.∑

ak(x− 1)k convergent at x = 3 =⇒ ∑ak(x− 1)k is absolutely convergent on (−1, 3).

(a)∑

ak =∑

ak(2 − 1)k; absolutely convergent

(b)∑

(−1)kak =∑

ak(0 − 1)k; absolutely convergent

(c)∑

(−1)kak2k =∑

ak(−1 − 1)k; ??

42. It must converge absolutely for −8 < x < 4.

43. (a) Suppose that∑

akrk is absolutely convergent. Then,

∑∣∣ak(−r)k∣∣ =∑ |ak|

∣∣(−r)k∣∣ =∑ |ak|

∣∣rk∣∣ =∑∣∣ak(−r)k∣∣ .

Therefore,∑∣∣ak(−r)k

∣∣ is absolutely convergent.

(b) If∑∣∣akrk∣∣ converged, then, from part (a),

∑∣∣ak(−r)k∣∣ would also converge.

44.∞∑

k=0

xk

rk

45.∞∑

k=0

= a0 + a1x + a2x2 + a0x

3 + a1x4 + a2x

5 + a0x6 + · · ·

=∞∑

k=0

(a0 + a1x + a2x

2)x3k

=∞∑

k=0

(a0 + a1x + a2x

2)(x3)k

(a)

∣∣∣∣∣(a0 + a1x + a2x

2)(x3)k+1

(a0 + a1x + a2x2) (x3)k

∣∣∣∣∣ = ∣∣x3∣∣ =⇒ r = 1.

(b)∞∑

k=0

(a0 + a1x + a2x

2)(x3)k =

(a0 + a1x + a2x

2) ∞∑k=0

(x3)k =(a0 + a1x + a2x

2) 1

1 − x3.

46. (−1, 1); sk ≤ k and∑

kxk converges for |x| < 1; for |x| ≥ 1, skxk �→ 0.

47. Examine the convergence of∑ |akxk|; for (a) use the root test and for (b) use the ratio rest.

48. By ratio test:∣∣∣∣ak+1

ak

∣∣∣∣ · |x| → |x|r, so

∣∣∣∣ak+1x2(k+1)

akx2k

∣∣∣∣ =∣∣∣∣ak+1

ak

∣∣∣∣|x|2 → |x|2r

=⇒ radius of convergence is√r.



SECTION 12.9 669

SECTION 12.9

1. Use the fact thatd

dx

(1

1 − x

)=

1(1 − x)2

:

1(1 − x)2

=d

dx(1 + x + x2 + x3 + · · · + xn + · · ·) = 1 + 2x + 3x2 + · · · + nxn−1 + · · · .

2.1

(1 − x)3=

12

d2

dx2

[1

1 − x

]=

12

d2

dx2

[1 + x + x2 + · · · + xn + · · ·

]=

12[2 + 6x + 12x2 + · · · + n(n− 1)xn−2 + · · ·

]= 1 + 3x + 6x2 + · · · + n(n− 1)

2xn−2 + · · ·

3. Use the fact thatd(k−1)

dx(k−1)

[1

1 − x

]=

(k − 1)!(1 − x)k

:

1(1 − x)k

=1

(k − 1)!d(k−1)

dx(k−1)

[1 + x + · · · + xk−1 + xk + xk+1 + · · · + xn+k−1 + · · ·

]=

1(k − 1)!

d(k−1)

dx(k−1)

[xk−1 + xk + xk+1 + · · · + xn+k−1 + · · ·

]= 1 + kx +

(k + 1)k2

x2 + · · · + (n + k − 1)(n + k − 2) · · · (n + 1)(k − 1)!

xn + · · ·

= 1 + kx +(k + 1)k

2!x2 + . . . +

(n + k − 1)!n!(k − 1)!

xn + · · · .

4. ln(1 − x) = −∫

dx

1 − x= −

[x +

x2

2+

x3

3+ · · · + xn+1

n + 1+ · · ·

]+ C; ln 1 = 0 =⇒ C = 0

=⇒ ln(1 − x) = −x− x2

2− x3

3− · · · − xn+1

n + 1− · · ·

5. Use the fact thatd

dx[ ln (1 − x2)] =

−2x1 − x2

:

11 − x2

= 1 + x2 + x4 + · · · + x2n + · · ·

−2x1 − x2

= −2x− 2x3 − 2x5 − · · · − 2x2n+1 − · · · .

By integration

ln (1 − x2) =(−x2 − 1

2x4 − 1

3x6 − · · · − x2n+2

n + 1− · · ·

)+ C.

At x = 0, both ln(1 − x2) and the series are 0. Thus, C = 0 and

ln (1 − x2) = −x2 − 12x4 − 1

3x6 − · · · − 1

n + 1x2n+2 − · · · .

6. ln(2 − 3x) = ln 2 + ln(1 − 3

2x)

= ln 2 − 3

2x− 1

2

(3

2

)2

x2 − 1

3

(3

2

)3

x3 − · · · − 1

n + 1

(3

2

)n+1

xn+1 − · · ·



670 SECTION 12.9

7. sec2 x =d

dx(tanx) =

d

dx

(x +

13x3 +

215

x5 +17315

x7 + · · ·)

= 1 + x2 +23x4 +

1745

x6 + · · ·

8. ln cosx = −∫

sinx

cosxdx = −

∫tanx dx = −x2

2− x4

12− x6

45− 17

2520x8 − · · · + C

ln cos 0 = 0 =⇒ C = 0

9. On its interval of convergence a power series is the Taylor series of its sum. Thus,

f(x) = x2 sin2 x = x2

(x− x3

3!+

x5

5!− x7

7!+ · · ·

)

= x3 − x5

3!+

x7

5!− x9

7!+ · · · =

∞∑n=0

f (n)(0)xn

n!

implies f (9)(0) = −9!/7! = −72.

10. f(x) = x cosx2 = x

(1 − (x2)2

2!+

(x2)4

4!− (x2)6

6!+ · · ·

)

=⇒ f (9)(0)9!

x9 =x9

4!=⇒ f (9)(0) =

9!4!

= 15120.

11. sinx2 =∞∑

k=0

(−1)k(x2)2k+1

(2k + 1)!=

∞∑k=0

(−1)kx4k+2

(2k + 1)!

12. x2 arctan x = x2

∫1

1 + x2dx = x2

∫ ( ∞∑k=0

(−1)kx2k

)dx = x2

[ ∞∑k=0

(−1)kx2k+1

2k + 1+ C

]

=∞∑

k=0

(−1)k

2k + 1x2k+3 (arctan 0 = 0 =⇒ C = 0)

13. e3x3=

∞∑k=0

(3x3)k

k!=

∞∑k=0

3k

k!x3k

14.1 − x

1 + x=

11 + x

− x

1 + x=

∞∑k=0

(−1)kxk −∞∑

k=0

(−1)kxk+1 = 1 + 2∞∑

k=0

(−1)k+1xk+1

15.2x

1 − x2= 2x

(1

1 − x2

)= 2x

∞∑k=0

(x2)k =∞∑

k=0

2x2k+1

16. x sinhx2 =x

2

(ex

2 − e−x2)

=x

2

[ ∞∑k=0

x2k

k!−

∞∑k=0

(−1)kx2k

k!

]=

x

2

[2

∞∑k=0

x4k+2

(2k + 1)!

]

=∞∑

k=0

x4k+3

(2k + 1)!



SECTION 12.9 671

17.1

1 − x+ ex =

∞∑k=0

xk +∞∑

k=0

xk

k!=

∞∑k=0

(k! + 1)k!

xk

18. coshx sinhx =12

sinh 2x =12

∞∑k=0

(2x)2k+1

(2k + 1)!=

∞∑k=0

4k

(2k + 1)!x2k+1

19. x ln (1 + x3) = x

∞∑k=1

(−1)k+1

k(x3)k =

∞∑k=1

(−1)k+1

kx3k+1

(12.6.8)∧

20. (x2 + x) ln(1 + x) = (x2 + x)∞∑

k=1

(−1)k+1

kxk = x2 +

∞∑k=3

(−1)k+1

(k − 1)(k − 2)xk

21. x3e−x3= x3

∞∑k=0

(−x3)k

k!=

∞∑k=0

(−1)k

k!x3k+3

22. x5(sinx + cos 2x) = x5

[ ∞∑k=0

(−1)k

(2k + 1)!x2k+1 +

∞∑k=0

(−1)k

(2k)!22kx2k

]

=∞∑

k=0

(−1)k

(2k + 1)![(2k + 1)4kx2k+5 + x2k+6

]

23. (a) limx→0

1 − cosxx2

=� limx→0

sinx

2x=

12

(� indicates differentiation of numerator and denominator).

(b) limx→0

1 − cosxx2

= limx→0

x2

2!− x4

4!+

x6

6!− · · ·

x2= lim

x→0

(12− x2

4!+

x4

6!− · · ·

)=

12

24. (a) limx→0

sinx− x

x2=� lim

x→0

cosx− 12x

=� limx→0

− sinx

2= 0

(b)sinx− x

x2=

−x3

3!+

x5

5!− · · ·

x2→ 0 as x → 0

25. (a) limx→0

cosx− 1x sinx

=� limx→0

− sinx

sinx + x cosx=� lim

x→0

− cosx2 cosx− x sinx

= −12

(b)

limx→0

cosx− 1x sinx

=−x2

2!+

x4

4!− x6

6!+ · · ·

x2 − x4

3!+

x6

5!· · ·

=− 1

2!+

x2

4!− x4

6!+ · · ·

1 − x2

3!+

x4

5!· · ·

= −12



672 SECTION 12.9

26. (a) limx→0

ex − 1 − x

x arctan x=� lim

x→0

ex − 1arctan x + x/(1 + x2)

=� limx→0

ex

11 + x2

+1 − x2

(1 + x2)2

=12

(b)ex − 1 − x

x arctan x=

x2

2+

x3

3!+

x4

4!+ · · ·

x2 − x4

3+

x6

5− · · ·

→ 12.

27.∫ x

0

ln(1 + t)t

dt =∫ x

0

1t

( ∞∑k=1

(−1)k−1

ktk

)dt =

∫ x

0

( ∞∑k=1

(−1)k−1

ktk−1

)dt

=∞∑

k=1

(−1)k−1

k

∫ x

0

tk−1 dt =∞∑

k=1

(−1)k−1

k2xk, −1 ≤ x ≤ 1

28.∫ x

0

1 − cos tt2

dt =∫ x

0

1t2

[ ∞∑k=1

(−1)k+1

(2k!)t2k

]dt

=∫ x

0

[ ∞∑k=1

(−1)k+1

(2k)!t2k−2

]dt =

∞∑k=1

(−1)k+1

(2k)!· x2k−1

2k − 1

29.∫ x

0

arctan t

tdt =

∫ x

0

1t

( ∞∑k=0

(−1)k

2k + 1t2k+1

)dt =

∫ x

0

( ∞∑k=0

(−1)k

2k + 1t2k

)dt

=∞∑

k=0

(−1)k

2k + 1

∫ x

0

t2k dt

=∞∑

k=0

(−1)k

(2k + 1)2x2k+1, −1 ≤ x ≤ 1

30.∫ x

0

sinh t

tdt =

∫ x

0

1t

[ ∞∑k=0

t2k+1

(2k + 1)!

]dt =

∫ x

0

∞∑k=0

t2k

(2k + 1)!dt =

∞∑k=0

x2k+1

(2k + 1)2(2k)!

31. 0.804 ≤ I ≤ 0.808; I =∫ 1

0

(1 − x3 +

x6

2!− x9

3!+ · · ·

)dx

=[x− x4

4+

x7

14− x10

60+

x13

(13)(24)− · · ·

]10

= 1 − 14

+114

− 160

+1

311− · · · .

Since1

311< 0.01, we can stop there:

1 − 14

+114

− 160

≤ I ≤ 1 − 14

+114

− 160

+1

311gives 0.804 ≤ I ≤ 0.808.



SECTION 12.9 673

32. I =∫ 1

0

(x2 − x6

3!+

x10

5!− x14

7!+ · · ·

)dx =

13− 1

7(3!)+

111(5!)

− 115(7!)

+ · · · .

Since1

11(5!)=

1840

< 0.01, we can stop there:

13− 1

7(3!)≤ I ≤ 1

3− 1

7(3!)+

111(5!)

gives 0.309 ≤ I ≤ 0.311.

33. 0.600 ≤ I ≤ 0.603; I =∫ 1

0

(x1/2 − x3/2

3!+

x5/2

5!− · · ·

)dx

=[23x3/2 − 1

15x5/2 +

1420

x7/2 − · · ·]10

=23− 1

15+

1420

− · · · .

Since1


23− 1

15≤ I ≤ 2

3− 1

15+

1420

gives 0.600 ≤ I ≤ 0.603.

34. I =∫ 1

0

(x4 − x6 +

x8

2!− x10

3!+

x11

4!− · · ·

)dx =

15− 1

7+

118

− 166

+1

288− · · · .

Since1


15− 1

7+

118

− 166

≤ I ≤ 15− 1

7+

118

− 166

+1

288gives 0.097 ≤ I ≤ 0.101.

35. 0.294 ≤ I ≤ 0.304; I =∫ 1

0

(x2 − x6

3+

x10

5− x14

7+ · · ·

)dx

(12.9.6)∧

=[13x

3 − 121x

7 + 155x

11 − 1105x

15 + · · ·]10

= 13 − 1

21 + 155 − 1

105 + · · · .

Since 1105 < 0.01, we can stop there:

13 − 1

21 + 155 − 1

105 ≤ I ≤ 13 − 1

21 + 155 gives 0.294 ≤ I ≤ 0.304.

36. I =∫ 2

1

(x

2!− x3

4!+

x5

6!− x7

8!+ · · ·

)dx

=3

2(2!)− 15

4(4!)+

636(6!)

− 2558(8!)

+ · · · =34− 15

96+

634320

− 255322560

+ · · ·.

Since255


34− 15

96+

634320

− 255322560

≤ I ≤ 34− 15

96+

634320

gives 0.607 ≤ I ≤ 0.609.



674 SECTION 12.9

37. I ∼= 0.9461; I =∫ 1

0

(1 − x2

3!+

x4

5!− · · ·

)dx

=[x− x3

3 · 3!+

x5

5 · 5!− · · ·

]10

= 1 − 13 · 3!

+1

5 · 5!− 1

7 · 7!· · · .

Since1

7 · 7!=

135, 280

∼= 0.000028 < 0.0001, we can stop there:

1 − 13 · 3!

+1

5 · 5!− 1

7 · 7!< I < 1 − 1

3 · 3!+

15 · 5!

; I ∼= 0.9461

38. I =∫ 0.5

0

(12!

− x2

4!+

x4

6!− x7

8!+ · · ·

)dx =

0.52!

− (0.5)3

3 · 4!+

(0.5)5

5 · 6!− (0.5)7

7 · 8!+ · · ·

(0.5)5

5 · 6!< 0.0001; I ∼= 0.2483

39. I ∼= 0.4485; I =∫ 0.5

0

(1 − x

2+

x2

3− x3

4+ · · ·

)dx

=[x− x2

22+

x3

32− x4

42+ · · ·

]1/20

=12− 1

22 · 22+

132 · 23

− 142 · 24

+ · · · =∞∑

k=1

(−1)k−1

k2 · 2k

Now,1

82 · 28=

116, 384

∼= 0.000061 is the first term which is less than 0.0001. Thus

7∑k=1

(−1)k−1

k2 · 2k < I <

8∑k=1

(−1)k−1

k2 · 2k ; I ∼= 0.4485

40. I =∫ 0.2

0

(x2 − x4

3!+

x6

5!− x8

7!+ · · ·

)dx =

(0.2)3

3− (0.2)5

5 · 3!+

(0.2)7

7 · 5!− · · ·

(0.2)5

5 · 3!< 0.0001, so I ∼= 0.0027

41. ex3; by (12.6.5) 42

∞∑k=0

1k!x3k+1 = x

∞∑k=0

1k!

(x3)k = xex3

43. 3x2ex3

=d

dx(ex

3)

44. (a) f(x) =ex − 1

x=

1x

(x +

x2

2!+

x3

3!+ · · ·

)= 1 +

x

2!+

x2

3!+

x3

4!+ · · ·

(b) f ′(x) =xex − ex + 1

x2=

12

+2x3!

+3x2

4!+ · · · + nxn−1

(n + 1)!+ · · ·

f ′(1) = 1 =∞∑

k=1

k

(k + 1)!



SECTION 12.9 675

45. (a) f(x) = xex = x

∞∑k=0

xk

k!=

∞∑k=0

xk+1

k!

(b) Using integration by parts:∫ 1

0

xex dx = [xex − ex]10 = e− e + 1 = 1.

Using the power series representation:∫ 1

0

xex dx =∫ 1

0

( ∞∑k=0

xk+1

k!

)dx =

∞∑k=0

∫ 1

0

(xk+1

k!

)dx

=∞∑

k=0

1k!

[xk+2

k + 2

]10

=∞∑

k=0

1k!(k + 2)

=12

+∞∑

k=1

1k!(k + 2)

Thus, 1 =12

+∞∑

k=1

1k!(k + 2)

and∞∑

k=1

1k!(k + 2)

=12.

46.d

dx(sinhx) =

d

dx

[ ∞∑k=0

x2k+1

(2k + 1)!

]=

∞∑k=0

(2k + 1)x2k

(2k + 1)!=

∞∑k=0

x2k

(2k)!= coshx

d

dx(coshx) =

d

dx

[ ∞∑k=0

x2k

(2k)!

]=

∞∑k=1

2kx2k−1

(2k)!=

∞∑k=1

x2k−1

(2k − 1)!= sinhx.

47. Let f(x) be the sum of these series; ak and bk are bothf (k)(0)

k!.

48. As k → ∞,

k1/k|x|(k−1)/k → |x|.

Thus, for k sufficiently large,

k1/k|x|(k−1)/k < |x| + ε and |kxk−1| = k|x|k−1 < (|x| + ε)k .

49. (a) If f is even, then the odd ordered derivatives f (2k−1), k = 1, 2, . . . are odd. This implies

that f (2k−1)(0) = 0 for all k and so a2k−1 = f (2k−1)(0)/(2k − 1)! = 0 for all k.

(b) If f is odd, then all the even ordered derivatives f (2k), k = 1, 2, . . . are odd. This implies

that f (2k)(0) = 0 for all k and so a2k = f (2k)(0)/(2k)! = 0 for all k.

50. f(0) = 1, f ′(0) = −2f(0) = −2, f ′′(x) = −2f ′(x) = 4f(x), f ′′(0) = 4

f (n)(x) = (−2)nf(x), f (n)(0) = (−2)n, f(x) =∞∑

k=0

(−2)k

k!xk =

∞∑k=0

(−2x)k

k!= e−2x



676 SECTION 12.9

51. f ′′(x) = −2f(x); f(0) = 0, f ′(0) = 1

f ′′(x) = −2f(x); f ′′(0) = 0

f ′′′(x) = −2f ′(x); f ′′′(0) = −2

f (4)(x) = −2f ′′(x); f (4)(0) = 0

f (5)(x) = −2f ′′′(x); f (5)(0) = 4

f (6)(x) = −2f (4)(x); f (6)(0) = 0

f (7)(x) = −2f (5)(x); f (7)(0) = −8...

f(x) = x− 23!x3 +

45!x5 − 8

7!x7 + · · · =

∞∑k=0

(−1)k 2k

(2k + 1)!x2k+1 =

1√2

sin(x√

2)

52. (a) f(x) = xe−x2 ln 2 = x− ln 2x3 +(ln 2)2

2x5 − (ln 2)3

6x7 +

(ln 2)4

24x9 + · · · .

f ′(x) = 1 − 3 ln 2x2 +5(ln 2)2

2x4 − 7(ln 2)3

6x6 +

3(ln 2)4

8x8 + · · · .

∫f(x) dx =

12x2 − ln 2

4x4 +

(ln 2)2

12x6 − (ln 2)3

48x8 +

(ln 2)4

240x10 + · · · .

(b) f(x) = x arctan x = x2 − 13 x

4 + 15 x

6 − 17 x

8 + 19 x

10 · · · .

f ′(x) = 2x− 43 x

3 + 65 x

5 − 87 x

7 + 109 x9 · · · .∫

f(x) dx =13x3 − 1

15x5 +

135

x7 − 163

x9 +199

x11 · · · .

53. 0.0352 ≤ I ≤ 0.0359; I =∫ 1/2

0

(x2 − x3

2+

x4

3− x5

4+ · · ·

)dx

=[x3

3− x4

8+

x5

15− x6

24+ · · ·

]1/20

=1

3(23)− 1

8(24)+

115(25)

− 124(26)

+ · · · .

Since1

24(26)=

11536


13(23)

− 18(24)

+1

15(25)− 1

24(26)≤ I ≤ 1

3(23)− 1

8(24)+

115(25)

gives 0.0352 ≤ I ≤ 0.0359. Direct integration gives

I =∫ 1/2

0

x ln (1 + x) dx =[12(x2 − 1) ln (1 + x) − 1

4x2 +

12x

]1/20

=316

− 38

ln 1.5 ∼= 0.0354505.



SECTION 12.9 677

54. I =∫ 1

0

(x2 − x4

3!+

x6

5!− x8

7!+ · · ·

)=

13− 1

5(3!)+

17(5!)

− 19(7!)

+ · · · .

Since1

9(7!)=

15040


13− 1

5(3!)+

17(5!)

− 19(7!)

≤ I ≤ 13− 1

5(3!)+

17(5!)

gives 0.3009 ≤ I ≤ 0.3011.

Direct integration gives

I =∫ 1

0

x sinx dx = [−x cosx + sinx]10 = sin 1 − cos 1 ∼= 0.3011686.

55. 0.2640 ≤ I ≤ 0.2643;

I =∫ 1

0

(x− x2 +

x3

2!− x4

3!+

x5

4!− x6

5!+

x7

6!− · · ·

)dx

=[x2

2− x3

3+

x4

4(2!)− x5

5(3!)+

x6

6(4!)− x7

7(5!)+

x8

8(6!)− · · ·

]10

=12− 1

3+

14(2!)

− 15(3!)

+1

6(4!)− 1

7(5!)+

18(6!)

− · · · .

Note that1

8(6!)=

15760

< 0.001. The integral lies between

12− 1

3+

14(2!)

− 15(3!)

+1

6(4!)− 1

7(5!)and

12− 1

3+

14(2!)

− 15(3!)

+1

6(4!)− 1

7(5!)+

18(6!)

.

The first sum is greater than 0.2640 and the second sum is less than 0.2643.

Direct integration gives∫ 1

0

xe−x dx =[−xe−x − e−x

]10

= 1 − 2/e ∼= 0.2642411.

56. For x ∈ [0, 4]

0 ≤ ex −(

1 + x +x2

2!+ · · · + xn

n!

)≤ e44n+1

(n + 1)!

(12.6.3)∧

Thus for x ∈ [0, 2]

0 ≤ ex2 −(

1 + x2 +x4

2!+ · · · + x2n

n!

)≤ e44n+1

(n + 1)!

It follows that

0 ≤∫ 2

0

ex2dx−

∫ 2

0

(1 + x2 +

x4

2!+ · · · + x2n

n!

)dx ≤

∫ 2

0

e44n+1

(n + 1)!dx

0 ≤∫ 2

0

ex2dx−

[x +

x3

3+

x5

5(2!)+ · · · + x2n+1

(2n + 1)n!

]≤ 2e44n+1

(n + 1)!

0 ≤∫ 2

0

ex2dx−

(2 +

23

3+

25

5(2!)+ · · · + 22n+1

(2n + 1)n!

)≤ e422n+3

(n + 1)!



678 SECTION 12.9

PROJECT 12.9A

1. f(x) = (1 + x)α f(0) = 1

f ′(x) = α(1 + x)α−1 f ′(0) = α

f ′′(x) = α(α− 1)(1 + x)α−2 f ′′(0) = α(α− 1)

and so on

f(x) = 1 + αx +α(α− 1)

2!x2 +

α(α− 1)(α− 2)3!

x3 + · · · .

2. ∣∣∣∣∣∣∣∣α(α− 1)(α− 2) · · · (α− k)

(k + 1)!α(α− 1)(α− 2) · · · (α− [k + 1])

k!

∣∣∣∣∣∣∣∣ =∣∣∣α−kk+1

∣∣∣→ 1 as k → ∞.

3. φ(x) = 1 + αx + α(α−1)2! x2 + α(α−1)(α−2)

3! x3 + · · · φ′(x) = α + α(α− 1)x + 12α(α− 1)(α− 2)x2 + · · ·

(1 + x)φ′(x) = φ′(x) + xφ′(x)

= α + α(α− 1)x +12α(α− 1)(α− 2)x2 + · · · + αx + α(α− 1)x2 +

12α(α− 1)(α− 2)x3 + · · ·

= α + α2x +α2(α− 1)

2!x2 +

α3(α− 1)(α− 2)3!

x3 + · · · = αφ(x)

4. g(x) =φ(x)

(1 + x)α; g′(x) =

(1 + x)αφ′(x) − αφ(x)(1 + x)α−1

(1 + x)2α=

(1 + x)φ′(x) − αφ(x)(1 + x)α+1

= 0.

Therefore, g(x) ≡ C, constant. Since g(0) = 1, C = 1; φ(x) = (1 + x)α.

5. (a) Take α = 1/2 in (12.9.7) to obtain 1 +12x− 1

8x2 +

116

x3 − 5128

x4.

(b)√

1 − x = [1 + (−x)]1/2 = 1 − x

2+

12 (− 1

2 )2!

x2 −12 (− 1

2 )(− 32 )

3!x3 +

12 (− 1

2 )(− 32 )(− 5

2 )4!

x4

= 1 − 12x− 1

8x2 − 1

16x3 − 5

128x4 − · · ·

(c) Replace x by x2 and take α = 1/2 to obtain√

1 − x2 ∼= 1 + 12x

2 − 18x

4.

(d) Replace x by −x2 and take α = 1/2:√

1 − x2 ∼= 1 − x2

2− 1

8x4

(e) Take α = −1/2 : 1 − 12x +

38x2 − 5

16x3 +

35128

x4.

(f) Take α = −1/4 :1

4√

1 + x= 1 − 1

4x +

532

x2 − 15128

x3 +1952048

x4 + · · ·



SECTION 12.9 679

6. (a) f(x) =1√

1 − x2=(1 − x2

)−1/2

In 12.9.7, replace x by x2 and take α = −1/2 to obtain

1√1 − x2

= 1 − 12x2 +

(−1/2)(−3/2)2!

x4 − (−1/2)(−3/2)(−5/2)3!

x6 + · · ·

= 1 − 12x2 +

38x4 +

516

x6 + · · ·

By Problem 2, this series has radius of convergence r = 1.

(b) arcsin x =∫ x

0

1√1 − x2

dt =∫ x

0

(1 − 1

2t2 +

38t4 +

516

t6 + · · ·)dt

= x− 16x3 +

340

x5 +5

112x7 + · · ·

By Theorem 12.9.3, the radius of convergence of this series is r = 1.

7. (a) α = −12

:1√

1 + x2= 1 − x2

2+

(− 12 )(− 3

2 )2!

x4 +(− 1

2 )(− 32 )(− 5

2 )3!

x6 + · · ·

= 1 − 12x2 +

38x4 − 5

16x6 + · · ·

(b) sinh−1 x =∫ x

0

1√1 + t2

dt =∫ x

0

(1 − 1

2t2 +

38t4 − 5

16t6 + · · ·

)dt

= x− 16x3 +

340

x4 − 5112

x7 + · · · ; r = 1

PROJECT 12.9B

1. tan [2 arctan (15 )] =

2 tan [arctan (15 )]

1 − tan2 [arctan (15 )]

=25

1 − 125

=512

2 arctan (15 ) = arctan ( 5

12 )

4 arctan (15 ) = 2 arctan ( 5

12 )

tan ([arctan (15 )] = tan [2 arctan (1

5 )] =1012

1 − 25144

=120119

tan [4 arctan (15 ) − arctan ( 1

239 )] =120119 − 1

239

1 + 120119 · 1

239

=120(239) − 119119(239) + 120

= 1

Thus 4 arctan (15 ) − arctan ( 1

239 ) =π

4.

2. 4 arctan (15 ) − arctan ( 1

239 ) < 45∑

k=1

(−1)k−1

2k − 1(15)2k−1 − [

1239

− 13(

1239

)3]

= 0.789582246 − 0.004184076 = 0.78539817.

4 arctan 15 − arctan 1

239 > 45∑

k=1

(−1)k−1

2k − 1(15)2k−1 − 1

239

= 0.789582238 − 0.0041841 = 0.785398138.

These inequalities imply 3.14159255 < π < 3.14159268.



680 REVIEW EXERCISES

3. 4 arctan 15 − arctan 1

239 < 415∑k=1

(−1)k−1

2k − 1

(15

)2k−1

−[

4∑k=1

(−1)k−1

2k − 1

(1

239

)2k−1]

= 0.785398163397448309616

4 arctan 15 − arctan 1

239 > 414∑k=1

(−1)k−1

2k − 1

(15

)2k−1[

3∑k=1

(−1)k−1

2k − 1

(1

239

)2k−1]

= 0.785398163397448306408

These inequalities imply 3.14159265358979322563 < π < 3.14159265358979323846.

REVIEW EXERCISES

1.∞∑

k=0

(34

)k

=1

1 − 34

= 4, a geometric series with r = 34 .

2.∞∑

k=0

(−1)k(12)k =

∞∑0

(−1

2

)k

=1

1 −(− 1

2

) =23, a geometric series with r = − 1

2

3. Since ex =∞∑

k=0

xk

k!,

∞∑k=0

(ln 2)k

k!= eln 2 = 2

4.∞∑

k=1

1k(k + 1)

=∞∑

k=1

(1k− 1

k + 1

)= lim

n→∞

(1 − 1

n + 1

)= 1


k


k2

7. converges; root test:(k + 13k

)1/k

→ 13

as k → ∞, or ratio test:k + 23k+1

3k

k + 1→ 1

3as k → ∞

8. diverges; ratio test:

(k + 1)!(k + 1)(k+1)/2

· kk/2

k!=

kk/2

(k + 1)(k−1)/2=(

k

k + 1

)k/2 √k + 1 → ∞.

9. converges; limit comparison with:∑ 1

k2

10. converges; root test:

[k

(34

)k]1/k

→ 34

as k → ∞

11. converges; ratio test,ak+1

ak=(k + 1k

)e

· 1e→ 1

e< 1

12. diverges; ratio test,ak+1

ak=

[2(k + 1)]!2k+1(k + 1)!

· 2kk!(2k)!

= 2k + 1 → ∞



REVIEW EXERCISES 681

13. converges; basic comparison,∑ (arctan k)2

1 + k2≤ π2

4

∑ 11 + k2

≤ π2

4

∑ 1k2

14. converges;2k + k4

3k=

2k

3k+

k4

3k, and each of the series

∑ 2k

3kand

∑ k4

3kis convergent.

15. absolutely convergent; basic comparison∑∣∣∣∣ (−1)k

(k + 1)(k + 2)

∣∣∣∣ ≤∑ 1k2

16. conditionally convergent:∑ (−1)k

2k + 1converges by Theorem 11.4.3;

∑∣∣∣∣ (−1)k

2k + 1

∣∣∣∣ =∑ 12k + 1

diverges.

17. absolutely convergent;∞∑

k=0

∣∣∣∣ (−1)k(100)k

k!

∣∣∣∣ = ∞∑k=0

100k

k!which converges by the ratio test.

18. conditionally convergent:∑ (−1)k√

(k + 1)(k + 2)converges by Theorem 11.4.3;

∑∣∣∣∣∣ (−1)k√(k + 1)(k + 2)

∣∣∣∣∣ =∑ 1√(k + 1)(k + 2)

diverges – limit comparison with∑ 1

k.

19. converges conditionally; Theorem 12.5.3:∑ ln k√

kdiverges by the integral test.

20. absolutely convergent; limit comparison with∑ 1

k3/2


k:

1k− 1

k + 1− 1

k + 2=

2 − k2

k(k + 1)(k + 2)

22. 1 − 122

+13− 1

42+ · · · =

∞∑k=0

(1

2k + 1− 1

(2k + 2)2

)=

∞∑k=0

4k2 + 6k + 3(2k + 1)(2k + 2)2

diverges; limit comparison with∑ 1

k.

23. ex =∞∑

k=0

xk

k!. Therefore,

xe2x2= x

∞∑k=0

(2x2)k

k!=

∞∑k=0

2k

k!x2k+1

24. ln(1 + x) =∞∑

k=1

(−1)k+1

kxk. Therefore,

ln(1 + x2) =∞∑

k=1

(−1)k+1

kx2k




25. arctan x =∞∑

k=0

(−1)kx2k+1

2k + 1. Therefore,

√x arctan

(√x)

= x1/2∞∑0

(−1)kx(2k+1)/2

2k + 1= x

12

∞∑0

(−1)kxk+ 12

2k + 1=

∞∑0

(−1)kxk+1

2k + 1

26. ax = ex lna. Therefore,

ax =∞∑

k=0

(ln a)k

k!xk

27. ln(1 + x) =∞∑

k=1

(−1)k+1

kxk. Therefore,

ln(1 + x2) =∞∑

k=1

(−1)k+1

k(x2)k and ln(1 − x2) =

∞∑k=1

(−1)k+1

k(−x2)k.

f(x) = x ln1 + x2

1 − x2= x[ln(1 + x2) − ln(1 − x2)] = x

( ∞∑k=1

(−1)k+1

k(x2)k −

∞∑k=1

(−1)k+1

k(−x2)k

)

= 2∞∑

k=0

x4k+3

2k + 1

28. sinx =∞∑

k=0

(−1)k

(2k + 1)!x2k+1. Therefore,

(x + x2) sinx2 = (x + x2)∞∑

k=0

(−1)k

(2k + 1)!(x2)2k+1 =

∞∑k=0

(−1)k

(2k + 1)!(x4k+3 + x4k+4

).

29. f(x) = (1 − x)1/3 f(0) = 1

f ′(x) = − 13 (1 − x)−

23 f ′(0) = − 1

3

f ′′(x) = − 29 (1 − x)−

53 f ′′(0) = − 2

9

f ′′′(x) = − 1027 (1 − x)−

73 f ′′′(0) = − 10

27

P3(x) = 1 − 13x− 1

9x2 − 5

81x3

30. f(x) = arcsin x f(0) = 0

f ′(x) =1√

1 − x2f ′(0) = 1

f ′′(x) =x

(1 − x2)3/2f ′′(0) = 0

f ′′′(x) =3x2

(1 − x2)5/2+

1(1 − x2)3/2

f ′′′(0) = 1

f (4)(x) =15x3

(1 − x2)7/2+

9x(1 − x2)5/2

f (4)(0) = 0

P4(x) = 1 + 16 x

3




31.[− 1

5 ,15

); ratio test:

bk+1

bk=

5kk + 1

|x| → 5|x| =⇒ r =15

At x = − 15 ,

∑ (−1)k

kconverges; at x = 1

5 ,∑ 1

kdiverges.

32. (−3, 3); ratio test:bk+1

bk=

13|x| =⇒ r = 3

at x = −3,∑ (−1)k

3k(−3)k+1 =

∑3(−1)2k+1 diverges;

at x = 3,∑ (−1)k

3k3k+1 =

∑3(−1)k diverges


bk=

2|x− 1|2(2k + 2)(2k + 1)

→ 0 =⇒ r = ∞

34. (0, 4); ratio test:bk+1

bk=

12|x− 2| =⇒ r = 2

at x = 0,∑ 1

2k(−2)k =

∑(−1)k diverges;

at x = 4,∑ 1

2k2k =

∑1 diverges

35. (−9, 9); ratio test:bk+1

bk=

k + 19k

|x| → 19|x| =⇒ r = 9

At x = −9,∑

k diverges; at x = 9,∑

(−1)kk diverges

36. (−1, 1); ratio test:bk+1

bk=

(k + 1)(2k + 1)k(2k + 3)

|x|2 → |x|2 =⇒ r = 1

at x = 1,∑ k

2k + 1diverges

(k

2k+1 → 12 as k → ∞

);

at x = −1,∑ −k

2k + 1diverges for the same reason.

37. (−4,−2]; ratio test:bk+1

bk=

√k√

k + 1|x + 3| → |x + 3| =⇒ r = 1

at x = −2,∑ (−1)k√

kconverges;

at x = −4,∑ 1√

kdiverges

38. diverges except at x = −1:bk+1

bk= (k + 1)|x + 1| → ∞ =⇒ r = 0

39. f(x) = e−2x = e−2(x+1)+2 = e2 · e−2(x+1) = e2∞∑0

[−2(x + 1)]k

k!= e2

∞∑0

(−1)k 2k

k!(x + 1)k; r = ∞.




40. sin 2x =∞∑0

(−1)k22k

(2k)!

(x− π

4

)2k

; r = ∞.

41. f(x) = lnx = ln[1 + (x− 1)] =∞∑1

(−1)k+1

k(x− 1)k; r = 1

42.√x + 1 = 1 +

12x− 1

8x2 +

∞∑3

(−1)k+1 1 · 3 · 5 · · · · · (2k − 3)2k k!

xk, r = 1

43.1

1 + x4=

∞∑k=0

(−1)kx4k

∫ 1/2

0

11 + x4

dx =∞∑

k=0

(−1)k∫ 1/2

0

x4kdx =∞∑

k=0

(−1)k1

4k + 11

24k+1

This is an alternating series with decreasing terms and the third term1

9(29)≈ 0.0002. Hence

∫ 1/2

0

11 + x4

dx ≈ 12− 1

5(25)≈ 0.4938

44. ex =n∑

k=0

xk

k!with remainder |Rn(x)| ≤ max |f (n+1)(t)| |x|

n+1

(n + 1)!.

Rn(2/3) < 3(2/3)n+1

(n + 1)!< 0.01 iff

(32

)n+1

(n + 1)! > 300 =⇒ n = 4.

Therfore e2/3 ≈ 1 + 23 + 1

2 (2/3)2 + 16 (2/3)3 + 1

24 (2/3)4 ≈ 1.9465

45. cosx =∞∑

k=0

(−1)k

(2k)!x2k;

1 − cos x

x2=

∞∑k=1

(−1)k+1

(2k)!x2(k−1)

∫ 1

0

1 − cosxx2

dx =∞∑

k=1

(−1)k+1

(2k)!1

2k − 1

This is an alternating series with decreasing terms and the 4th term1

8!(7)< 0.0001. Therefore

∫ 1

0

1 − cosxx2

dx ≈3∑

k=1

(−1)k+1

(2k)!1

2k − 1≈ 0.4864

46. sinx =∞∑

k=0

(−1)k

(2k + 1)!x2k+1; x sin x4 =

∞∑k=0

(−1)k+1

(2k + 1)!x8k+5.

∫ 1

0

x sinx4 dx =∞∑

k=0

(−1)k

(2k + 1)!1

8k + 6





5!(24)< 0.01. Therefore

∫ 1

0

x sinx4 dx ≈ 16− 1

3!(14)≈ 0.155

47. Let g(x) = sinx and a = π/4. Then sin x =√

22

+√

22

(x− π/4) −√

22(2!)

(x− π/4)2 −√

22(3!)

(x− π/4)3 + · · ·∣∣Rn(x)

∣∣ = ∣∣g(n+1)(c)∣∣

(n + 1)!

∣∣ (x− π

4

)n+1 ∣∣≤∣∣ (x− π/4)

∣∣n+1

(n + 1)!(g(n+1)(c) = ± sin c or ± cos c)

Now, 48◦ =48π180

radians. We want to find the smallest positive integer n such that∣∣Rn(48π/180 − π/4)∣∣ < 0.0001.

∣∣Rn(48π/180 − π/4)∣∣ ≤ ( π

60

)n+1 1(n + 1)!

∼= (0.05236)n+1

(n + 1)!< 0.0001 =⇒ n ≥ 2.

sin x ∼= P2(x) =√

22

+√

22

(x− π

4

)−

√2

4

(x− π

4

)2

; sin 48◦ ∼=√

22

[1 +

π

60− 1

2

( π

60

)2]∼= 0.7432

48.∫ 1

0

x2e−x2dx =

∞∑k=0

∫ 1

0

(−1)k

k!x2k+2 dx =

∞∑k=0

(−1)k

k!1

2k + 3


5!(13)< 0.001. Therefore

∫ 1

0

x2e−x2dx ≈

4∑k=0

(−1)k

k!1

2k + 3≈ 0.1900

49. For the sine function, x− 16 x

3 + 1120 x

5 = P5 = P6. Therefore, for x ∈ [0, π/4] we have

| sinx− P5(x)| =∣∣∣∣f (7)(c)

7!x7

∣∣∣∣ ≤ 17!

(π4

)7

< 0.000037

(∣∣f (7)(c)∣∣ = cos c ≤ 1

) ∧

50. For the cosine function, 1 − 12 x

2 + 124 x

4 − 1720 x

6 = P6 = P7. Therefore, for x ∈ [0, π/4] we have

| cosx− P6(x)| =∣∣∣∣f (8)(c)

8!x8

∣∣∣∣ ≤ 18!

(π4

)8

< 0.0000036

(∣∣f (8)(c)∣∣ = cos c ≤ 1

) ∧




51.∞∑

k=1

ak =∫ ∞

1

xe−x dx =2e

52. Let ε > 0. For each positive integer n, set an = xn − ε

2n+1and bn = xn +

ε

2n+1. Then

bn − an =ε

2nand

∞∑n=1

(bn − an) =∞∑

n=1

ε

2n=

ε

2< ε

53. If∞∑

k=1

(ak+1 − ak) converges, then the sequence of partial sums sn = an+1 − a1 converges.

Therefore, the sequence ak converges.

If the sequence ak converges, then the sequence sn = an+1 − a1 converges which implies that the

series∞∑

k=1

(ak+1 − ak) converges.

54. (a) ak =∞∑

n=0

(1k

)n =1

1 − 1/k=

k

k − 1and

∞∑k=2

ak =∞∑

k=2

k

k − 1.

The series diverges because ak =k

k − 1�→ 0.

(b) ak =∞∑

n=1

(1k

)n =1/k

1 − 1/k=

1k − 1

and∞∑

k=2

ak =∞∑

k=2

1k − 1

.

The series diverges; limit comparison with∑ 1

k.

(c) ak =∞∑

n=2

(1k

)n =1/k2

1 − 1/k=

1k(k − 1)

and∞∑

k=2

ak =∞∑

k=2

1k(k − 1)

.

The series converges; limit comparison with∑ 1

k2.

calculus one and several variables 10e salas solutions manual ch12

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