summary of convergence tests for series and solved problems integral test ratio test root test...

Summary of Convergence Tests for Series and Solved ProblemsIntegral Test

Ratio TestRoot TestComparison Theorem for SeriesAlternating Series

Mika Seppälä: Series

Test Test Quantity

Converges if

Diverges if

Ratio q < 1 q > 1

Root r < 1 r > 1

Integral Int < ∞ Int = ∞

q =limk→ ∞

ak+1

ak

r =lim

k→ ∞a

kk

Int = f x( )dx

1

∞

∫

S = a

kk=1

∞

∑In the Integral Test we assume that there is a decreasing non-negative function f such that ak = f(k) for all k. The Test Quantity of the Integral Test is the improper integral of this function.

The above test quantities can be used to study the convergence of the series S.


Comparison Test and the Alternating Series Test

Comparison Test

bk

k=1

∞

∑ converges then also the series ak

k=1

∞

∑ converges, and ak

k=1

∞

∑ ≤ bk

k=1

∞

∑ .

Conversely, if ak

k=1

∞

∑ diverges, then also the series bk

k=1

∞

∑ diverges.

Assume that 0≤ ak ≤ bk for all k. If the series

The alternating series (−1)k+1akk=1

∞

∑ =a1 −a2 +a3 −K ,

converges if

Alternating Series Test

1

2 lim

k→ ∞ak =0.

∀k :ak ≥ak+1 ≥0, and


Error EstimatesError Estimate by the Integral Test

Let ak=f a

k( ) where f is a decreasing positive function. I f ak

k=1

∞

∑

converges by the Integral Test, then ak

k=1

M

∑ − ak

k=1

∞

∑ ≤ f x( )dxM

∞

∫ .

If the alternating series (−1)k+1ak1

∞∑ converges by the

Alternating Series Test, then (−1)k+1ak1

M∑ − (−1)k+1ak1

∞∑ ≤aM+1.

Error Estimates by the Alternating Series Test

Error of the approximation by the Mth partial sum.

This means that the error when estimating the sum of a converging alternating series is at most the absolute value of the first term left out.


Overview of Problems

2

Assume that ak and b

k are positive for all k, a

kk=1

∞

∑ converges

and that limk→ ∞

bk

ak

=L < ∞. Show that the series bk

k=1

∞

∑ converges.

1

Let ak and b

k be positive for all k. Assume that the series a

kk=1

∞

∑

and bk

k=1

∞

∑ both converge. Show that the series ak

k=1

∞

∑ bk converges.

3

We know that ak4k converges.

k=1

∞

∑ Do the following

series a) ak

−4( )k

k=1

∞

∑ and b) ak2 k

k=1

∞

∑ converge?


Overview of Problems

6

Compute 2

k k + 2( )k=1

∞

∑

7

k2

k3 +1k=1

∞

∑ 8

1

k3 + kk=1

∞

∑ 9

10 Estimate

1

k2

k=1

∞

∑ with error < 0 .001

Do the above series 4-5 and 7 – 9 converge or diverge?

Show that if a

k> 0 and lim

k→ ∞ka

k> 0, then a

kk=1

∞

∑ diverges.11

12 Show that if a

k> 0 and lim

k→ ∞k2a

k=1, then a

kk=1

∞

∑ converges.

4

1

k ln k( )k=2

∞

∑ 5

sin

1

k

⎛

⎝⎜

⎞

⎠⎟

k=1

∞

∑


Overview of Problems13

−1( )k+1

kk=1

∞

∑ 14

−1( )k+1

k

ln k( )k=1

∞

∑ 15

−1( )k+1

k + k( )

k − kk=2

∞

∑

16

17

Estimate

−1( )k+1

k3k=1

∞

∑ with error < 0 .001 .

18

Estimate −1( )

k

2k +1( )!k=0

∞

∑ with error < 0 .0001 .

19

−1( )k k2

2 kk=1

∞

∑ 20

−1( )k 2 k

k2k=1

∞

∑

Do the series in 13 – 16 converge?

Do the series in 19 – 20 converge?


Overview of Problems21

For which values of the parameter p the series

1

k pk=1

∞

∑ converges?

22

For which values of the parameter p the series 1

k lnp k( )k=2

∞

∑ converges?

23

k −22k + 3

⎛

⎝⎜

⎞

⎠⎟

k

k=1

∞

∑ 24

k20

k !k=1

∞

∑ 25

26

e−n n!n=1

∞

∑ 27

n2 e−n

n=1

∞

∑

28 Let F

n be the nth Fibonacci number.

Fn

2nn=1

∞

∑

29 Let a

1=1, a

n+1=

sin n( ) + cos n( )

na

n. a

nn=1

∞

∑

30 Show that the series

xn

n!

n=0

∞

∑ converges for all values of x.

Do the series given in Problems 23 – 29 converge?


Comparison Test

1

Let ak and b

k be positive for all k. Assume that

the series ak

k=1

∞

∑ and bk

k=1

∞

∑ both converge. Show

that the series ak

k=1

∞

∑ bk converges.Solution

Therefore there is a number m

1 such that k > m

1⇒ b

k< 1 .

Now k > m

1⇒ a

kb

k< a

k. Recall that by the assumptions a

kb

k> 0 .

Remark: it suffices to show that akb

km1

∞∑ converges because

akb

k1

∞∑ = akb

k1

m1 −1∑ + akb

km1

∞∑ and the sum akb

k1

m1 −1∑ is finite.

The Comparison Theorem now implies that a

kb

km1

∞∑ converges.


Comparison Test

2

Let ak and b

k be positive for all k. Assume that

the series ak

k=1

∞

∑ converges and that limk→ ∞

bk

ak

=L < ∞.

Show that the series bk

k=1

∞

∑ converges.Solution

Since lim

k→ ∞

bk

ak

=L, there is a number m1 such that k > m

1⇒

bk

ak

≤L +1 .

Therefore k > m

1⇒ b

k≤ L +1( )ak

.

Since a

k1

∞∑ converges, also L +1( )ak1

∞∑ converges.

Remark: it suffices to show that bkm

1

∞∑ converges because

bk1

∞∑ = bk1

m1 −1∑ + bkm1

∞∑ and the sum bk1

m1 −1∑ is finite.

The Comparison Theorem now implies that b

km1

∞∑ converges.


The Comparison Test

3


k=1

∞


series a) ak

−4( )k

k=1

∞

∑ and b) ak2 k

k=1

∞

∑ converge?

Solution

I f ak= −1( )

k 1k4 k

then ak4 k

k=1

∞

∑ =−1( )

k

kk=1

∞

∑ converges by the

Alternating Series Test but ak

−4( )k

k=1

∞

∑ =1kk=1

∞

∑ is the Harmonic

Series which diverges.

The series a) needs not converge.

Example:


1

Hence the series

2 converges absolutely.kk

k k

a

The Comparison Test3


k=1

∞


series a) ak

−4( )k

k=1

∞

∑ and b) ak2 k

k=1

∞

∑ converge?

Solution (cont’d)

Also k > k

1⇒ a

k2 k = a

k4 k 2 k

4 k

The series b) does converge.

Since ak4k

k=1

∞

∑ converges,

limk→ ∞

ak4 k =0.

The Geometric Series 1

2

⎛

⎝⎜

⎞

⎠⎟

k

k=k1

∞

∑ converges.

We conclude, by the Comparison Test,

that the series ak2k

k=1

∞

∑ converges.

<

12

⎛

⎝⎜

⎞

⎠⎟

k

.

<1


The Integral Test

4

Study the convergence of 1

k ln k( ).

k=2

∞

∑

Solution

The series 1

k ln k( )2

∞∑ diverges by the Integral Test

since dx

x ln x( )2

∞

∫ = limM→ ∞

dx

x ln x( )2

M

∫ = limM→ ∞

ln ln x( )( )⎤⎦2

M

= limM→ ∞

ln ln M( )( ) −ln ln 2( )( )( ) =∞.


Comparison Test

5 Study the convergence of sin

1

k

⎛

⎝⎜

⎞

⎠⎟ .

k=1

∞

∑

Solution We know that

x

2≤sin x( ) for 0 ≤x ≤1.

Therefore

1

2k≤sin

1k

⎛

⎝⎜

⎞

⎠⎟ for k ≥1.

The series

1

2k1

∞∑ diverges by the Integral Test

since dx

2x1

∞

∫ = limM→ ∞

dx2 x1

M

∫ = limM→ ∞

ln x( )

2

⎤

⎦

⎥⎥1

M

= limM→ ∞

ln M( )

2=∞.

The Comparison Theorem now implies that sin

1

k

⎛

⎝⎜

⎞

⎠⎟1

∞∑ diverges.

From Applications of Differentiation.


Partial Fraction Computation

6

Solution

Use the Partial Fraction Decomposition 2

k k + 2( )=1k−

1k + 2

.

Sm=

2

k k + 2( )k=1

m

∑ =1k−

1k + 2

⎛

⎝⎜

⎞

⎠⎟

k=1

m

∑

=

1kk=1

m

∑ −1

k + 2k=1

m

∑ = 1 +12+13+L +

1m

⎛

⎝⎜

⎞

⎠⎟ −

13+L +

1m

+1

m +1+

1m + 2

⎛

⎝⎜

⎞

⎠⎟

These terms cancel.


Comparison Test

7

Study the convergence of

k2

k3 +1.

k=1

∞

∑

Solution Observe that for k ≥1, 2k3 ≥k3 +1 .

Since the Harmonic Series 1

kk=1

∞

∑ diverges,

also the series 12kk=1

∞

∑ diverges.

By the Comparison Test, the series

k2

k3 +1k=1

∞

∑ diverges.


The Integral and the Comparison Tests

8

Study the convergence of

1

k3 + k.

k=1

∞

∑

Solution

Observe first that, by the Integral Test, 1

k3k=1

∞

∑ converges since

dx

x31

∞

∫ = limM→ ∞

dx

x31

M

∫ = limM→ ∞

−1

2 x21

M

= limM→ ∞

−1

2M2− −

12

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜

⎞

⎠⎟ =

12 converges.

Since 0 ≤1

k3 + k≤

1k3

, the series 1

k3 + kk=1

∞

∑ converges

by the Comparison Test.


The Integral Test

9

Study the convergence of 1

k ln k( ) ln ln k( )( ).

k=4

∞

∑

Solution

To use the Integral Test we have to compute

dx

x ln x( ) ln ln x( )( )4

∞

∫ = limM→ ∞

dx

x ln x( ) ln ln x( )( ).

4

M

∫

To compute dx

x ln x( ) ln ln x( )( )4

M

∫ use the substitution t =ln ln x( )( ).

Then dt =dx

x ln x( ) and we get

dx

x ln x( ) ln ln x( )( )4

M

∫ =dtt

ln ln 2( )( )

ln ln M( )( )

∫ =ln ln ln M( )( )( ) −ln ln ln 2( )( )( )M→ ∞ → ∞.

Hence the series diverges by the Integral Test.


The Integral Test10

Show that 1

k2

k=1

∞

∑ converges and

estimate the sum with error < 0 .001 Solution

The convergence follows from the fact that

dx

x21

∞

∫ = limM→ ∞

dx

x21

M

∫ = limM→ ∞

−1x

⎛

⎝⎜

⎞

⎠⎟

1

M

= limM→ ∞

−1M

− −1( )⎛

⎝⎜

⎞

⎠⎟ =1.

To find out how many terms we need to take in our approxiation,

determine M so that dx

x2M

∞

∫ < 0 .001 .

Computing as before,

dx

x2M

∞

∫ =1M

< 0 .001 if M > 1000 .

Computing 1000th partial sum by Maple we get the approximation 1.6439. The precise value of the above infinite sum is π2/6≈1.6449.


The Comparison Test

Show that if a

k> 0 and lim

k→ ∞ka

k> 0, then a

kk=1

∞

∑ diverges.

Solution Let lim

k→ ∞ka

k=b > 0 .

By the definition of limits: ∃m

1 such that k > m

1⇒ ka

k>

b2.

Hence, k > m

1⇒ a

k>

b2k

.

The series b

2kk=m1

∞

∑ , diverges since the Harmonic Series diverges.

Hence, by the Comparison Test, a

kk=1

∞

∑ diverges.

You can show this also directly by the Integral Test without referring to the Harmonic Series.

11



Show that if a

k> 0 and lim

k→ ∞k2a

k=1, then a

kk=1

∞

∑ converges.

Solution Since lim

k→ ∞k2a

k=1, ∃m

1 such that k > m

1⇒ k2a

k< 2 .

Hence: k > m

1⇒ 0 < a

k<

2k2

.

The series 2

k2k=m1

∞

∑ converges by the Integral Test.

This follows since the improper integral

2dx

x2m

1

∞

∫ = limM→ ∞

2dx

x2m1

M

∫ = limM→ ∞

−2x

m1

M

= limM→ ∞

−2M

− −1m

1

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

1m

1

converges.

Hence, by the Comparison Test, a

kk=1

∞

∑ converges.


The Alternating Series Test13

Does the series

−1( )k+1

kk=1

∞

∑ converge?

Solution

The series is of the form −1( )k+1

ak

k=1

∞

∑

where ak=

1

k.

Hence the sequence a

k( ) is decreasing and limk→ ∞

ak=0.

Hence, by the Alternating Series Test,

−1( )

k+1

kk=1

∞

∑ converges.



Does the series −1( )

k+1k

ln k( )k=1

∞

∑ converge?

Solution


ak

k=1

∞

∑

where ak=

k

ln k( ).

Since limk→ ∞

ak=lim

k→ ∞

k

ln k( )=∞,

the series −1( )

k+1k

ln k( )k=2

∞

∑ diverges.



Does the series

−1( )k+1

k + k( )

k − kk=2

∞

∑ converge?

Solution


ak

k=2

∞

∑

where ak=

k + k

k − k.

Since limk→ ∞

ak=lim

k→ ∞

k + k

k − k=lim

k→ ∞

1 +1

k

1 −1

k

=1 ≠0,

the series −1( )

k+1k + k( )

k − kk=2

∞

∑ diverges.




ksin

πk

⎛

⎝⎜

⎞

⎠⎟

k=2

∞

∑ converge?

Solution

The series is of the form −1( )k

ak

k=2

∞

∑

where ak=sin

πk

⎛

⎝⎜

⎞

⎠⎟ > 0 for all k ≥2.

Since limk→ ∞

ak=0 and since the sequence a

k( ) is decreasing,

the series −1( )ksin

πk

⎛

⎝⎜

⎞

⎠⎟

k=2

∞

∑ converges

by the Alternating Series Test.

This follows from the fact that the sine function is increasing for 0≤x≤π/2.



Estimate the sum

−1( )k+1

k3k=1

∞

∑ with error < 0 .001 .

Solution

The series is a convergent alternating seriesq

of the form −1( )k+1

ak

k=1

∞

∑ where ak=

1k3

.

The error of the approximation −1( )

k+1

k3k=1

m

∑ ≈−1( )

k+1

k3k=1

∞

∑

is ≤1

m +1( )3.

31

The error is 0.001 if 0.001, 1

i.e., if 1 10 or 9.

m

m m

Hence the desired estimate is

−1( )k+1

k3k=1

9

∑ ≈0.90211 .



Estimate the sum −1( )

k

2k +1( )!k=0

∞

∑ with error < 0 .0001 .

Solution

The series is a convergent alternating series

of the form −1( )k

ak

k=0

∞

∑ where ak=

1

2k +1( )!.

The error of the approximation −1( )

k

2k +1( )!k=0

m

∑ ≈−1( )

k

2k +1( )!k=0

∞

∑

is ≤1

2m + 3( )!.

1The error is 0.0001 if 0.0001,

2 3 !

i.e., if 3.

m

m

Hence the desired estimate is −1( )

k+1

2k +1( )!k=0

3

∑ =42415040

≈0.841468 .




k k2

2 kk=1

∞

∑ converge?

Solution


ak

k=1

∞

∑

where ak=

k2

2 k.

Since limk→ ∞

ak=lim

k→ ∞

k2

2 k=0,

the series −1( )k k2

2 kk=1

∞

∑ converges.

Use l’Hospital’s Rule.

19




k 2 k

k2k=1

∞

∑ converge?

Solution


ak

k=1

∞

∑

where ak=2 k

k2.

Since limk→ ∞

ak=lim

k→ ∞

2 k

k2=∞,

the series −1( )k 2 k

k2k=1

∞

∑ diverges.

Use l’Hospital’s Rule.


The Integral Test21

For which values of the parameter p

the series 1

k pk=1

∞

∑ converges?

Solution

The improper integral

dx

xp1

∞

∫ converges if and only if p > 1 .

Hence, by the Integral Test, 1

k pk=1

∞

∑ converges

if and only if p > 1 .


The Integral Test22

For which values of the parameter p

the series 1

k lnp k( )k=2

∞

∑ converges?

Solution

Hence, by the Integral Test, 1

k lnp k( )k=1

∞

∑ converges

if and only if p > 1 .

To compute dx

x lnp x( )1

M

∫ use the substitution t =ln x( ) , dt =dxx

.

=t−p+1

1 −pln2

lnM

=lnM( )

1−p

1 −p−

ln2( )1−p

1 −pM→ ∞ → −

ln2( )1−p

1 −p if 1 −p < 0 .

This requires that p≠1. If p=1, the corresponding improper integral diverges.


The Root Test23

Does the series

k −22k + 3

⎛

⎝⎜

⎞

⎠⎟

k

k=1

∞

∑ converge?

Solution

ak

k =k −22k + 3

⎛

⎝⎜

⎞

⎠⎟

k

k =k −22k + 3

=1 −

2k

2 +3k

k→ ∞ → 12

< 1 .

Hence

k −22k + 3

⎛

⎝⎜

⎞

⎠⎟

k

k=1

∞

∑ converges by the Root Test.

Use the Root Test.


The Ratio Test24

Does the series

k20

k !k=1

∞

∑ converge?

Solution

ak+1

ak

=

k +1( )20

k +1( )!

k20

k !

=k +1

k

⎛

⎝⎜

⎞

⎠⎟

20k !

k +1( )!= 1 +

1k

⎛

⎝⎜

⎞

⎠⎟

201

k +1k→ ∞ → 0 .

Hence

k20

k !k=1

∞

∑ converges by the Ratio Test.

Use the Ratio Test.



Does the series

cos k( ) + sin k( )

k2k=1

∞

∑ converge?

Solution

ak=

cos k( ) + sin k( )

k2<

2k2

for all k.

Use the Comparison Test.

Hence the series

cos k( ) + sin k( )

k2k=1

∞

∑ converges absolutely.

According to Problem 21.

Conclude that the series converges.


The Ratio Test26

Solution

an+1

an

=e

− n+1( ) n +1( )!

e−n n!=

en

en+1

n +1( )!

n!=1e

n +1( )n→ ∞ → ∞

Hence e−n n!

n=1

∞

∑ diverges by the Ratio Test.

Use the Ratio Test.


The Ratio Test27

Solution

an+1

an

=n +1( )

2e

− n+1( )

n2 e−n=

n +1n

⎛

⎝⎜

⎞

⎠⎟

2en

en+1= 1 +

1n

⎛

⎝⎜

⎞

⎠⎟

21e

n→ ∞ → 1e

Hence n2 e−n

n=1

∞

∑ converges.

Use the Ratio Test.


The Ratio Test

28

Let F0=0,F

1=1 and F

n=F

n−1+ F

n−2. The numbers F

n are the Fibonacci

numbers. They can be determined by the formula Fn=α n −βn

α −β where

α =1 + 5

2 and β =

1 − 52

.

Solution

Does the series

Fn

2nn=1

∞

∑ converge?

an+1

an

=

Fn+1

2 n+1

Fn

2 n

=F

n+1

Fn

2 n

2 n+1

=

α −βα

⎛

⎝⎜⎞

⎠⎟

n

β

1 −βα

⎛

⎝⎜⎞

⎠⎟

n

12

Conclude that the series converges by the Ratio Test.

=

α n+1 −βn+1

α −βα n −βn

α −β

12

=α n+1 −βn+1

α n −βn

12

n→ ∞ → α2

< 1 .

Now use the fact that β < α.

Hence 0 <βα

< 1 and limn→ ∞

βα

⎛

⎝⎜

⎞

⎠⎟

n

=0.


The Ratio Test

29

Let a1=1, a

n+1=

sin n( ) + cos n( )

na

n.

Does the series an

n=1

∞

∑ converge?

Solution

an+1

an

=

sin n( ) + cos n( )

na

n

an

=sin n( ) + cos n( )

nn→ ∞ → 0

The series converges by the Ratio Test.

Observe that for all positive integers n, sin(n) + cos(n) ≠0. Hence, for every n, an≠ 0, and the above ratio is defined for all n.


The Ratio Test30

Show that the series

xn

n!

n=0

∞

∑ converges for all values of x.

Solution

an+1

an

=

xn+1

n +1( )!

xn

n!

=n!

n +1( )!

xn+1

xn=

x

n +1n→ ∞ → 0 .

Hence

xn

n!n=0

∞

∑ converges regardless of the value of x.

Use the Ratio Test.

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