math 116 — practice for exam 2mconger/dhsp/116w18exam2... · math 116 — practice for exam 2...

Math 116 — Practice for Exam 2

Generated March 18, 2018

Name: SOLUTIONS

Instructor: Section Number:

1. This exam has 15 questions. Note that the problems are not of equal difficulty, so you may want toskip over and return to a problem on which you are stuck.

2. Do not separate the pages of the exam. If any pages do become separated, write your name on themand point them out to your instructor when you hand in the exam.

3. Please read the instructions for each individual exercise carefully. One of the skills being tested onthis exam is your ability to interpret questions, so instructors will not answer questions about examproblems during the exam.

4. Show an appropriate amount of work (including appropriate explanation) for each exercise so that thegraders can see not only the answer but also how you obtained it. Include units in your answers whereappropriate.

5. You may use any calculator except a TI-92 (or other calculator with a full alphanumeric keypad).However, you must show work for any calculation which we have learned how to do in this course. Youare also allowed two sides of a 3′′ × 5′′ note card.

6. If you use graphs or tables to obtain an answer, be certain to include an explanation and sketch of thegraph, and to write out the entries of the table that you use.

7. You must use the methods learned in this course to solve all problems.

Semester Exam Problem Name Points Score

Winter 2011 2 7 rod gravity 7

Fall 2006 1 8 instruction speed 12

Winter 2012 2 9 wire 14

Winter 2013 3 4 tree 10

Fall 2007 2 5 16

Fall 2009 2 6 camp 9

Fall 2017 2 7 bouncy ball 12

Winter 2003 3 9 nautilus 6

Fall 2003 3 8 pendulum 12

Winter 2015 3 11 checkers 12

Fall 2008 2 6 15

Fall 2008 2 8 16

Fall 2015 3 13 10

Fall 2013 3 7 8

Fall 2004 2 3 10

Total 169

Recommended time (based on points): 170 minutes

Math 116 / Exam 2 (March 2011) page 10

7. [7 points] A rod of length L meters has mass density δ0, where 0 ≤ x ≤ L represents theposition in meters along the rod measured from its left endpoint. The force of gravitationalattraction F between the rod and a particle of mass m lying in the same line as the rod at adistance a is given by

F =

∫L

0

Gmδ0

(a+ x)2dx.

where G is the constant of gravitation.

In certain cases (when the mass of the particle is small and the rod is long), you can assumethat the rod has infinite length. Calculate the gravitational force between a rod of infinitelength and a particle of mass m which is a meters away (arranged as shown above).

Solution:

∫∞

0

Gmδ0

(a+ x)2dx = Gmδ0 lim

b→∞

∫b

0

1

(a+ x)2dx

= Gmδ0 limb→∞

−1

(a+ x)|b0

= Gmδ0 limb→∞

−1

(a+ b)+

1

a

=Gmδ0

a

University of Michigan Department of Mathematics Winter, 2011 Math 116 Exam 2 Problem 7 (rod gravity) Solution

Math 116 / Exam 1 (October 11, 2006) page 9

8. [12 points] In class, Chris’ calculus professor is well known to cover material at a rate m(t) =1

12(t−20)2/3 textbook sections/minute, where t is the time in minutes since the start of class.

(a) [2 of 12 points] What is the meaning of the integral∫ 80

0m(t) dt (include units in your explanation)?

Solution:The integral is the area under the rate m(t) between t = 0 and t = 80, which, by the FundamentalTheorem of Calculus is the total number of sections that Chris’ professor covers in the 80 minute(90 less ten) class period.

(b) [4 of 12 points] How many sections would you estimate the professor covers in the first minute ofclass? In the 20th minute? Why?

Solution:We note that m(0) = 1

12(20)2/3 = 0.0113 sections/minute. Thus we might guess that the pro-

fessor might cover approximately 0.0113 sections-worth of material in the first minute. If we re-peat this calculation for the 20th minute, we might guess that the number of sections covered is(1 minute)(m(19)) = 1

12(1)2/3 = 112 sections. Note, however, that m(20) is undefined—therefore, it

appears that the professor is speaking at an infinite rate at about t = 20, so that we might alsowonder if an infinite number of words are spoken. We can verify this by completing part (c) of thisproblem.

(c) [6 of 12 points] Find exactly (that is, by hand) the value of∫ 80

0m(t) dt.

Solution:We note that because m(t) is discontinuous at t = 20 this is an improper integral. We thereforeevaluate

∫ 80

0

1

12(t − 20)2/3dt = lim

a→20−

∫ a

0

1

12(t− 20)2/3dt + lim

a→20+

∫ 80

a

1

12(t− 20)2/3dt

= lima→20−

1

4(t − 20)1/3

∣

∣

∣

∣

a

0

+ lima→20+

1

4(t − 20)1/3

∣

∣

∣

∣

80

a

= lima→20−

1

4

(

(a − 20)1/3 + (20)1/3)

+ lima→20+

1

4

(

(60)1/3− (a − 20)1/3

)

=1

4(20)1/3 +

1

4(60)1/3 sections.

Or, approximately 1.66 sections per class period. Because this integral is finite, it is clear that theamount of material covered in the 20th minute is, in fact, finite too.

University of Michigan Department of Mathematics Fall, 2006 Math 116 Exam 1 Problem 8 (instruction speed) Solution

Math 116 / Exam 2 (March 19, 2012) page 10

9. [14 points] A machine produces copper wire, and occasionally there is a flaw at some pointalong the wire. The length x of wire produced between two consecutive flaws is a continuousvariable with probability density function

f(x) =

{

c(1 + x)−3 for x ≥ 0

0 otherwise.

Show all your work in order to receive full credit.

a. [5 points] Find the value of c.

Solution: Since f(x) is a density function∫

∞

−∞f(x)dx = 1. Then

∫

∞

−∞

f(x)dx =

∫

∞

0c(1 + x)−3dx = lim

b→∞

∫

b

0c(1 + x)−3dx

= limb→∞

−c

2(1 + x)2

∣

∣

∣

b

0 =c

2= 1

Hence c = 2.

b. [3 points] Find the cumulative distribution function P (x) of the density function f(x).Be sure to indicate the value of P (x) for all values of x.

Solution:

P (x) =

∫

x

−∞

f(t)dt =

∫

x

0c(1 + t)−3dt =

−c

2(1 + t)2|x0 =

c

2−

c

2(1 + x)2= 1−

1

(1 + x)2.

P (x) =

{

1− 1(1+x)2

x ≥ 0.

0 x < 0.

University of Michigan Department of Mathematics Winter, 2012 Math 116 Exam 2 Problem 9 (wire) Solution

Math 116 / Exam 2 (March 19, 2012) page 11

c. [5 points] Find the mean length of wire between two consecutive flaws.

Solution:

mean =

∫

∞

−∞

xf(x)dx =

∫

∞

0c

x

(1 + x)3dx = c lim

b→∞

∫

b

0

x

(1 + x)3dx

u = x v′ = (1 + x)−3

u′ = 1 v =−1

2(1 + x)2

= c limb→∞

−x

2(1 + x)2

∣

∣

∣

b

0 +

∫

b

0

1

2(1 + x)2dx

= c limb→∞

−x

2(1 + x)2−

1

2(1 + x)

∣

∣

∣

b

0 =c

2= 1.

or

mean =

∫

∞

−∞

xf(x)dx =

∫

∞

0c

x

(1 + x)3dx = c lim

b→∞

∫

b

0

x

(1 + x)3dx

u = 1 + x

= c limb→∞

∫

b+1

1

u− 1

u3dx = c lim

b→∞

∫

b+1

1u−2 − u−3dy

= c limb→∞

−u−1 +u−2

2

∣

∣

∣

b+11 =

c

2= 1.

d. [1 point] A second machine produces the same type of wire, but with a different probabilitydensity function (pdf). Which of the following graphs could be the graph of the pdf forthe second machine? Circle all your answers.

Solution: The graph on the left upper corner can’t be the density sincex is the distancebetween flaws, hence the probability density function can’t be positive for x < 0.The graph on the left lower corner can’t be the density since the area under the curvefor x ≥ 0 is infinite (it has a positive horizontal asymptote).The graph on the right upper corner can’t be a density since it is negative on an interval.

x

y

x

y

x

y

University of Michigan Department of Mathematics Winter, 2012 Math 116 Exam 2 Problem 9 (wire) Solution

Math 116 / Final (April 26, 2013) page 7

4. [10 points] The lifetime t (in years) of a tree has probability density function

f(t) =

a

(t+ 1)pfor t ≥ 0.

0 for t < 0.

where a > 0 and p > 1.

a. [4 points] Use the comparison method to find the values of p for which the averagelifetime M is finite (M < ∞). Properly justify your answer.

Solution: The average lifetime M is given by the formula M =

∫

∞

0

ta

(t+ 1)pdt.

Sincet

a

(t+ 1)p≤ t

a

tp=

a

tp−1for t > 0,

then∫

∞

1

ta

(t+ 1)pdt ≤

∫

∞

1

a

tp−1dt

We know that a

∫

∞

1

1

tp−1converges precisely when p−1 > 1 ( p > 2) by the p-test,

so the first integral converges precisely when p > 2. This implies that the averagelifetime M is finite for p > 2.Note:

•We use the inequality

∫

∞

1

ta

(t+ 1)pdt ≤

∫

∞

1

a

tp−1dt since the inequality

∫

∞

0

ta

(t+ 1)p∫

∞

0

a

tp−1dt is not useful

(∫

∞

0

a

tp−1dt = ∞ for all values of p

)

.

•You do not need to discuss the convergence of the integral

∫

1

0

at

(t+ 1)pdt since

this integral is not an improper integral.

University of Michigan Department of Mathematics Winter, 2013 Math 116 Exam 3 Problem 4 (tree) Solution


b. [4 points] Find a formula for a in terms of p. Show all your work.

Solution: We know that

1 =

∫

∞

0

a

(t+ 1)pdt.

We use u-substition with u = t+ 1 to calculate the integral:

∫

∞

0

a

(t+ 1)pdt = lim

b→∞

∫ b

0

a

(t+ 1)pdt

= limb→∞

∫ b+1

1

a

updu = a lim

b→∞

∫ b+1

1

u−pdu

= a limb→∞

u−p+1

(−p+ 1)|b+1

1 = a limb→∞

1

(−p+ 1)up−1|b+1

1

(since p > 1) =a

p− 1.

Therefore 1 =a

p− 1, so a = p− 1.

c. [2 points] Let C(t) be the cumulative distribution function of f(t). For a given tree,what is the practical interpretation of the expression 1− C(30)?

Solution: 1− C(30) is the probability that a given tree lives at least 30 years.

University of Michigan Department of Mathematics Winter, 2013 Math 116 Exam 3 Problem 4 (tree) Solution

Math 116 / Exam 2 (November 14, 2007) page 6

t = 0 t = 10

b

a

5. [16 points] For all parts of this problem, refer to thegraph to the right, which gives a cumulative distri-bution function P (t) for some density function p(x).The given graph shows all important features of thedistribution (for values of t greater and less than thoseshown, the behavior shown continues).

(a) [4 points of 16] What are the y-values a and b?Why?

Solution:We know that the smallest value of the cdf is zero,so b = 0. Similarly, the largest value is one, soa = −1/5.

(b) [4 points of 16] What is the approximate value of the median of this distribution?

Solution:The median is where the cdf has the value 1

2 . This occurs at t ≈ 1.75.

(c) [4 points of 16] Suppose that two points on the graph are (3.9, 0.90) and (4.1, 0.92). Estimate p(4).

Solution:The density function p(x) is just the derivative of the cumulative distribution function P (t), so we

expect p(4) ≈ 0.92−0.94.1−3.9 = 0.10. Alternately, we know that

∫ 4.1

3.9p(x) dx = P (4.1)−P (3.9) = 0.02, so

because∫ 4.1

3.9p(x) dx ≈ p(4) · (0.2), we have p(4) · (0.2) ≈ 0.02, and therefore p(4) ≈ 0.10.

(d) [4 points of 16] Continue to suppose that two points on the graph are (3.9, 0.90) and (4.1, 0.92).

Estimate∫ 4

0p(x) dx.

Solution:Again, we know that

∫ 4

0p(x) dx = P (4), so

∫ 4

0p(x) dx ≈

12 (0.92 + 0.90) = 0.91.

University of Michigan Department of Mathematics Fall, 2007 Math 116 Exam 2 Problem 5 Solution


6. [9 points] Camp Summerama is a summer camp for teenagers. The camp is open for eightweeks every summer, and campers are able to attend for any length of time desired, between 0to 8 weeks. The function p(x) is the probability density function that the campers will enrollfor x number of weeks. It is a piecewise function, defined by

p(x) =

{

c

5x 0 ≤ x ≤ 5

− c

3x + 8c

35 < x ≤ 8

and shown in the graph below.

0 1 2 3 4 5 6 7 80

c

x

f1 f

2

(5,c)

a. [2 points] What is the value of c?

Solution: c = 1

4= 0.25

b. [3 points] Evaluate p(7). Interpret your answer in a complete sentence, using the contextof campers and weeks spent at camp.

Solution: Given that c = 1

4, we have p(7) = − 1

12(7) + 8

12= 1

12. For a small interval ∆x,

approximately 1

12∆x of the campers spent between 7 and 7 + ∆x weeks at camp.

c. [4 points] Determine the median value for this density function. Interpret your answer ina complete sentence, using the context of campers and weeks spent at camp.

Solution:

0.5 =

∫

T

0

p(x)dx =

∫

T

0

1

20xdx =

1

40x2|T0 =

1

40T 2

T =√

20 ≈ 4.472

Half of the campers spend less than 4.472 weeks at camp, and half spend more than 4.472weeks at camp.

University of Michigan Department of Mathematics Fall, 2009 Math 116 Exam 2 Problem 6 (camp) Solution


7. [12 points] A bouncy ball is launched up 20 feet from the floor and then begins bouncing.Each time the ball bounces up from floor, it bounces up again to a height that is 60% theheight of the previous bounce. (For example, when it bounces up from the floor after falling20 ft, the ball will bounce up to a height of 0.6(20) = 12 feet.)Consider the following sequences, defined for n ≥ 1:

• Let hn be the height, in feet, to which the ball rises when the ball leaves the ground forthe nth time. So h1 = 20 and h2 = 12

• Let fn be the total distance, in feet, that the ball has traveled (both up and down) whenit bounces on the ground for the nth time. For example, f1 = 40 and f2 = 40 + 24 = 64.

a. [2 points] Find the values of h3 and f3.

Solution: h3 = 0.6(12) = 7.2 and f3 = 64 + 14.4 = 78.4.

Answer: h3 = 7.2 and f3= 78.4

b. [6 points] Find a closed form expression for hn and fn.(“Closed form” here means that your answers should not include sigma notation or ellipses(· · · ). Your answers should also not involve recursive formulas.)

Solution: hn = 0.6hn−1 is a recursive relationship that holds between the terms of thesequence hn for n > 1, and this recursive formula means that hn is a geometric sequence.The (constant) ratio of successive terms is equal to 0.6 and first term is h1 = 20. So wesee that hn = 20(0.6)n−1.

Note that the term fn is twice the sum of the first n terms of the hn sequence. (Twicebecause the bouncy ball travels both up and down.) We use the formula for a partialsum of a geometric series (i.e. a finite geometric series) to find

fn = 2(h1 + h2 + ...+ hn) = 2(20 + ...+ 20(0.6)n−1)

=2(20)(1− (0.6)n)

1− 0.6=

40(1− (0.6)n)

0.4= 100(1− (0.6)n).

Answer: hn = 20 · (0.6)n−1and fn=

40(1−(0.6)n)0.4 = 100(1− (0.6)n)

University of Michigan Department of Mathematics Fall, 2017 Math 116 Exam 2 Problem 7 (bouncy ball) Solution


c. [4 points] Decide whether the given sequence or series converges or diverges.If it diverges, circle “diverges”. If it converges, circle “converges” and write the value towhich it converges in the blank.

i. The sequence fn

Converges to 100 Diverges

Solution: The limit of the sequence fn is

limn→∞

fn = limn→∞

40(1− (0.6)n)

0.4=

40

0.4= 100.

Since this limit exists, the sequence fn converges, and this computation shows that itconverges to 100.

Alternatively, as we saw in part b, the sequence fn is the sequence of partial sums of

the geometric series∞∑

k=1

2hk =∞∑

k=1

40(0.6)k−1. Since r = 0.6 and |0.6| < 1, we know that

this geometric series converges to40

1− 0.6= 100. By definition of series convergence, this

sum is the limit of the sequence of partial sums fn, i.e. limn→∞

fn = 100.

ii. The series

∞∑

n=1

hn

Converges to 50 Diverges

Solution: Next, we consider the series

∞∑

n=1

hn, which we know is geometric from part

b. Since the common ratio between successive terms is 0.6, the series converges, and theformula for the sum of a convergent geometric series gives us

∞∑

n=1

hn =

∞∑

n=1

20 · (0.6)n−1 =20

1− 0.6= 50,

Alternatively, since the sequence fn is the sequence of partial sums of the series∞∑

k=1

2hk,

we have

∞∑

n=1

hn =1

2limn→∞

fn =100

2= 50.

University of Michigan Department of Mathematics Fall, 2017 Math 116 Exam 2 Problem 7 (bouncy ball) Solution

9

9. (6 pts) The chambered nautilus builds a spiral sequence of closed chambers. It constructsthem from the inside out, with each chamber approximately

� ��larger (by volume) than the

last. (The large open section at the top is not a “chamber.”) The largest chamber is 9 cubicinches. How much volume is enclosed by all the chambers? Assumefor simplicity that there areinfinitely many chambers. Show your work.

Because the chambers grow by a constant factor each time, they form a geometric series. Ifeach is

� ��larger than the previous, then the ratio between them is� � � . But this is the ratio of

the larger divided by the smaller, and we want the opposite, sowe get � � � � � � � � �� . This isthe ratio by which you have to multiply each volume to get the next smaller volume. The totalvolume, then, is: � � � � �

� � � � �

� � � � � �

� � �!� �!�

This geometric series sums to��

. So the total enclosed volume is 54 cubic inches.

By the way, the numbers given in this problem are not simply made up, but are deduced fromthe size and shape of a large adult chambered nautilus. The number

� is the approximate volume

of a cylinder with height 2 inches and radius 3 inches (a rough approximation to the organism’ssize and shape).

Where does�� come from? Notice that one “band” of the chambers takes about� � cham-

bers, and (by directly measuring the picture), shrinks the organism by a factor of�, in length.

Scaling down by a factor of�

in length is the same as scaling by a factor of��

in volume, whichshould leave

� �� cubic inches. Therefore the first� � chambers take

��cubic inches. So

we have the equations:

�

� � ��

and� � � � � � �

� � ��

Solving simultaneously gives��

, � � � � �� . This is how the problem waswritten.

University of Michigan Department of Mathematics Winter, 2003 Math 116 Exam 3 Problem 9 (nautilus) Solution

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University of Michigan Department of Mathematics Fall, 2003 Math 116 Exam 3 Problem 8 (pendulum) Solution


11. [12 points] You construct a snowflake by starting with a square piece of paper of side length3 inches. You divide the square into a three by three grid of squares of side length one andremove the four squares in the grid that share a side with the center square in the grid. Foreach remaining square in the grid, subdivide each of them into 9 equally sized squares andremove the four squares in each of these new grids that share a side with the center square inthe grid. You continue in this manner for a long time.

N = 0 N = 1 N = 2 N = 3

a. [3 points] Write a formula that gives the perimeter, PN , of the black squares that makeup the snowflake after N steps.

Solution: PN = 12

(

5

3

)N

b. [2 points] Find limN→∞

PN .

Solution: PN tends to infinity as N → ∞.

c. [3 points] Suppose N ≥ 1. Write a sum that gives the area, AN of all the squares youhave removed after N steps.

Solution:

N−1∑

j=0

4

(

5

9

)j

d. [2 points] Write a closed form expression for AN .

Solution: AN = 41−

(

5

9

)N

1− 5

9

e. [2 points] Find the limit as N → ∞ of your expression in (d).

Solution: limN→∞

AN = 9

University of Michigan Department of Mathematics Winter, 2015 Math 116 Exam 3 Problem 11 (checkers) Solution


6. [15 points] For each of the following, assume that∑

an and∑

bn are both convergent series,and that an > an+1 > bn > bn+1 > 0. For each, explain your answer in a sentence or two, orwith a clear picture or counterexample.

a. [3 points] Is∑

(bn − an) a convergent series? Explain.

Solution: Yes,∑

(bn − an) is a convergent series. We know∑

an and∑

bn converge,so

∑(bn − an) must converge to

∑bn −

∑an.

b. [3 points] Is∑

(an · bn) a convergent series? Explain.

Solution: Yes,∑

an · bn is convergent. We know that an → 0 as n → ∞, so for somesufficiently large value of n, an < 1. Thus for sufficiently large values of n, an·bn < bn, andtherefore because

∑bn converges, by the comparison test

∑an · bn must also converge.

c. [3 points] Is∑

((−1)n ln(an + 1)) a convergent series? Explain.

Solution: Yes,∑

(−1)n ln(an + 1) is a convergent series. We know (because∑

an

converges) that an → 0 as n → ∞, so ln(an + 1) → ln(1) = 0, and we’re given thatan > an+1 > 0, so ln(an + 1) > ln(an+1 + 1) > 0. Thus ln(an + 1) → 0 monotonicallyfrom above as n → ∞, and

∑(−1)n ln(an + 1) is an alternating series. Thus, by the

alternating series test we know that the series converges.

d. [3 points] Is∑

(2 an) a convergent series? Explain.

Solution: Yes,∑

2 an is a convergent series. Because∑

an is convergent, 2∑

an isclearly well-defined, and 2

∑an =

∑2 an.

e. [3 points] Is∑

((−1)n

√bn ) an absolutely convergent series? Explain.

Solution: We can’t tell if∑

(−1)n√

bn is absolutely convergent. This requires that∑ √bn converge, which we can’t tell because for sufficiently large n it must be that

√bn > bn, and thus the convergence of

∑bn doesn’t tell us what happens to

∑ √bn.



8. [16 points] For each of the following series, state a convergence test that you could use todetermine if the series converges or not and indicate why you chose that test. Then carefullyapply the test to determine if the series converges or not. Mathematical precision is importantin this problem.

a. [8 points]∑

n=2

√n + 3

n2 − 1

Solution: This looks like a good problem to try and do a comparison test, but the most

obvious comparison (√

nn2 = 1

n3/2) isn’t larger than the given function, so we use the Limit

Comparison Test. We know that∑

n−3/2 converges. The limit we consider is lim

n→∞

anbn

,

which with an =√

n+3

n2−1and bn = 1

n3/2is

limn→∞

√n + 3

n2 − 1·n

3/2

1= lim

n→∞

n2

√

1 + 3

n

n2 − 1= lim

n→∞

√

1 + 3

n

1 − 1

n2

= 1.

This is a finite non-zero limit, so we know that the convergence properties of both ofthese series are the same, so the given series must converge.

b. [8 points]∑ (n + 1)!

2e3n

Solution: This problem involves factorials and exponentials, so the ratio test is a goodone to use. We have

limn→∞

∣

∣

∣

∣

an+1

an

∣

∣

∣

∣

= limn→∞

(

(n + 2)!

(2 e3n+3)!

)(

2e3n

(n + 1)!

)

= limn→∞

(

n + 2

e3

)

→ ∞.

This diverges, so by the ratio test the series diverges.


Math 116 / Final (December 17, 2015) DO NOT WRITE YOUR NAME ON THIS PAGE page 11

13. [10 points] Suppose an and bn are sequences with the following properties.

•

∞∑

n=1

an converges.

• n ≤ bn ≤ en.

For each of the following statements, decide whether the statement is always true, sometimestrue, or never true. Circle your answer. No justification is necessary. You only need to

answer 5 of the 7 questions. Only answer the 5 questions you want graded. If it is unclearwhich 5 questions are being answered, the first 5 questions you answer will be graded.

a. [2 points] The sequence1

bndiverges.

ALWAYS SOMETIMES NEVER

b. [2 points] The sequence an is bounded.


c. [2 points] The series

∞∑

n=1

1

bndiverges.


d. [2 points] The series

∞∑

n=1

e−an converges.


e. [2 points] The series∞∑

n=1

a2

ndiverges.


f. [2 points] The series∞∑

n=1

anbn converges.


g. [2 points] The series∞∑

n=1

bn

n!converges.



Math 116 / Final (December 17, 2013) page 10

7. [8 points] Consider the power series

∞∑

n=1

2n

3n(x− 5)n.

In the following questions, support your answers by stating and properly justifying any test(s),facts and computations you use to prove convergence or divergence. Show all your work.

a. [4 points] Find the radius of convergence of the power series.

Solution:

limn→∞

∣

∣

∣

∣

∣

∣

2n+1

3(n+1)(x− 5)n+1

2n

3n(x− 5)n

∣

∣

∣

∣

∣

∣

= |x− 5| limn→∞

2n

n+ 1= 2|x− 5| then

1

R= 2

or

R = limn→∞

∣

∣

∣

∣

∣

2n

3n2n+1

3(n+1)

∣

∣

∣

∣

∣

= limn→∞

n+ 1

2n=

1

2.

Radius of convergence=0.5

b. [4 points] Find the interval of convergence of the power series. Make sure to cite all thetests you use to find your answer.

Solution: Testing the endpoints:

•x = 4.5:∞∑

n=1

2n

3n(4.5− 5)n =

1

3

∞∑

n=1

(−1)n

nconverges by alternating series test.

•x = 5.5:∞∑

n=1

2n

3n(5.5− 5)n =

1

3

∞∑

n=1

1

ndiverges by p-series test p = 1 ≤ 1.

Interval of convergence: 4.5 ≤ x < 5.5.


5

3. (10 points)

(a) Find the radius of convergence R of the following power series. Show your work.

∞∑

n=1

(n + n3 2n)

n2 3n(x − 1)n .

The general coefficient of the given power series is an =n + n3 2n

n2 3n(x − 1)n. We need to find the

limit as n goes to infinity of the ratio∣

∣an+1/an

∣

∣. For this, observe that for large values of n, the

term an “behaves” liken3 2n

n2 3n(x − 1)n = n (2/3)n(x − 1)n. Thus we have:

limn→∞

∣

∣

∣

∣

an+1

an

∣

∣

∣

∣

= limn→∞

(n + 1)(2/3)n+1 |x − 1|n+1

n(2/3)n |x − 1|n=

2

3lim

n→∞

n + 1

n|x − 1| =

2

3|x − 1| .

Therefore, the radius of convergence of the power series is3

2.

(b) What is the interval of convergence of the series?

The power series is given around the point x = 1, and we have found its radius of convergence

to be 3/2. Accordingly, the series converges for values of x within the point 1 − 3/2 = −1/2 ,

and the point 1 + 3/2 = 5/2.

The interval of convergence is therefore: −1

2< x <

5

2.


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