math 116 — practice for exam 3math 116 / final (december 17, 2015) do not write your name on this...
TRANSCRIPT
Math 116 — Practice for Exam 3
Generated April 24, 2019
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
Fall 2015 3 5 3
Winter 2016 3 11 Hanoi Tower 5
Winter 2004 2 6 Okefenokee 12
Winter 2017 2 2 lemniscate 12
Winter 2011 3 5 wood 14
Fall 2010 2 7 fishtank 15
Winter 2011 2 4 grasshoppers2 12
Winter 2019 2 9 Infinity Tower 9
Fall 2016 3 11 Equinate 10
Winter 2017 3 8 sand dollar 8
Winter 2009 2 4 switchboard 50
Fall 2018 1 8 pond 12
Fall 2018 3 1 15
Fall 2016 2 9 10
Winter 2011 3 10 10
Total 197
Recommended time (based on points): 167 minutes
Math 116 / Final (December 17, 2015) DO NOT WRITE YOUR NAME ON THIS PAGE page 5
4. [8 points] Let f(x) = 3√
1 + 2x2.
a. [5 points] Find the first 3 nonzero terms of the Taylor series for f centered at x = 0.
Solution: Using the Taylor series for (1 + y)1/3 centered at y = 0,
3
√
1 + y = 1 +1
3y +
(
1
3
) (
−2
3
)
2!y2 + . . .
= 1 +y
3−
y2
9+ . . .
Substituting y = 2x2,
3√
1 + 2x2 = 1 +2x2
3−
4x4
9+ . . .
b. [3 points] For what values of x does the Taylor series converge?
Solution: The binomial series for 3√1 + y converges when −1 < y < 1. Substituting
y = 2x2, this converges when 1 < 2x2 < 1, or −1√2< x < 1√
2.
5. [3 points] Determine the exact value of the infinite series
−1 +1
3!−
1
5!+ · · ·+
(−1)n+1
(2n+ 1)!+ . . .
No decimal approximations are allowed. You do not need to show your work.
Solution:
−1 +1
3!−
1
5!+ · · · =
∞∑
n=0
(−1)n+1
(2n+ 1)!=
∞∑
n=0
(−1)n(−1)2n+1
(2n+ 1)!= sin(−1).
University of Michigan Department of Mathematics Fall, 2015 Math 116 Exam 3 Problem 5 Solution
Math 116 / Final (April 21, 2016) DO NOT WRITE YOUR NAME ON THIS PAGE page 12
11. [5 points] The Hanoi tower is made by rotating the region depicted below around the y-axis.The region is made up of infinitely many adjacent rectangles. The nth rectangle has width 1
and height an =1
n!(2n+ 1)where n = 0, 1, 2, 3, .... The rectangle touching the y-axis corre-
sponds to n = 0. Note that the y-axis is not to scale.
x1 2 3 4 5 6 7
y
a4
a3
a2
a1
a0
a5.. .
Compute the volume of the Hanoi Tower. Give an exact answer.
Solution: To compute the volume of the Hanoi Tower, we focus on each rectangle separately.The volume of the object made by the revolution of the nth rectangle is given by
[π(n+ 1)2 − πn2] · an = π(2n+ 1)
1
n!(2n+ 1)=
π
n!
The total volume is given by adding the volume of all those objects for n = 0, 1, 2, 3, ....
∞∑
n=0
π
n!= π ·
∞∑
n=0
1
n!= πe
University of Michigan Department of Mathematics Winter, 2016 Math 116 Exam 3 Problem 11 (Hanoi Tower) Solution
�k� � §p©��0���] ���� 0 %�*Y�� ��� d $�W$��R��� ��c��c��] %�('��*�% %�*� ��q �� �^Y d �W�$ \t��� O �('� �Y��� d �')���a %�P�?$R�'�YJ %�iR�'��*YJ ����� �Y�� O �*�&Z,$ �� W�$ % �W�� ')�*�% (` A � � d $�W$��R��� ��aYJ')�?�i���$ %��')�� R[Y��#� O �D��� d $')�����n� �:Z,� %�D�?$R�')YJ %�iYJ�g %�('�%�0�*������ R? �� � Z,��] %�('c��P ����vZ YJ')� O W���i�*�_���@ ���� F ���(�g�*� � ���(�?�%Z Y��i� ` A ��� W$� O YJ �$��q���V �� ����� O ��hk��s ��� �C�=������'��*��� ���V�?�W��*�;� ��'� ��khXY������ ��� �C�]���#� '��*���C���m�?�W$�*�C�*Y��% a���� �� � d Y���� O Y��i�q��� ���� d $�W���R��% ��(h� ��Y*\n�,YJ�g %�(', �� �*�';� �(�#YJ'� �� ')�c� '���. �� � F ���(�g�*� � ���(�v�%Z Y��i�P��CR$Q��*� d \s ��� �V<Jª � + §©�� ��� �@< �� $ + §¬ 8y%z0{ � � �('��v��C �� � F ���(�g�*� � ���(� �%Z Y��?�P���'��*W�YJ ����b %�? �� � d Y��%� O Y��i�P��� �� � d $�W$��R��% ��lS
0 � <�°nh �� � d $'����_YJ')�?YJ ! �� �?�%Z Y��i� h �%�t� &��cW$� O YJ %�*� s � °��&<68�¬��=������'��*� �?�W$�*�&� ��'� ��rY��#�� � °��@< 8�©!�=����� ')�*�?�?�W$�*�V�*Y��% ����0 �� � d Y��%� O Y��i�^���k ���� d $�W���R��% ��(` A ��YJ ��(h,¬�° �?�W$�*�@� ��'� ���Y��#�©�°?�?�W��*� �*Y��% (`y%}@{ 6 � �� �('��a�(Q��(',Yv ���i�aZ,� �*�[ �� ����� O �?��� d $')�#�@��� %')Y*Q��*W�W��� R ��� � / ��') ��n� 3 Y��% �S 6 � �%� h Z,� �*��S 6 �L� �� (h� U �#W�Y����Z,�]\�� �� (`A � � ��� O �9��� %')Y*Q��*W�W��� R � ��'� �� � �*Y��% tYJ b ���i� � $�!s#~ ��� �r<��=~ ��� � 5 °n` ? �� O �qs#~ ��� �r< ª Y��#�� ~ ��� �m< 21 � K � 1 h# ����� � O(O � ')�CZ,���*��ª!< 21 � K � 1 ��';Z,� �*� � < ¨?�#Y*\n�(`y��n{ 6 � �� ����� O ����� d $')��� O ����% �Y��] �W$\^�i�TQ=�� R �� '��� R� ��� � �� �a"�')�� � �� ')�(�&�#Y*\n� ��� �� �*$'��%�� ')� �(\�S � ��\��';Z,�]\�� �� �SA � �&Q��*W$� O $ \���� �� ����� O ��YJ ; ���?� � ���� ��� ��<�� � s ~ ��� �%� 1 + � � ~ ��� �%� 1 <�� ¯ +��x ��� ª 5 °nh#�%�i ������� O ���� O ����% �Y��] �W$\���t�i�� ����t��� ')�� R Y�W�W0 �� ����Y:\=�;� ;�� %')Y*Q��*W�W��� R `y��@{ � �:Z �oYJ',� �=�*�C �� �!��� O �^��� d �')��� %')Y*Q��*WL��� �� �!"#')�% �� '��(���#Y*\n�C���� �� �*$'"�%�� ')� �(\�S? �� O �@ �� �@')YJ %����� O �#Y�� R��������#��% �Y�� O �V %'�Y*Q��*W�W$�*� Z,� ��c'��*�%�0� O %�a ���i� �� ����@Q��*W$� O $ \�h� ���������� �Y�� O � %')Y*Q��*W�W$�*�t��� '��� R �� �!"�')�� ���'��(����Y:\=�;��CR$Q��*� d \� �� �!��] %�(R�')Y�W `# 2� � ��� �%$ � < # 2�'& ¯ + ¯
¨ � $ �F � � O Y�� O Y�W O ��W�YJ %�r �����P��� %�(R�'�Y�W h_�*� �� �(' d \9���)�� R �� �r�oY O P ���YJ tY�� Y��] ��� �('�$Q5YJ ��Q��r� ��'t ������] %�(R�')Y���� �� « � §c+)(x � 2 � 1 Y��#� YJ���#W$\n�� Rq �� �b�o������Y��?�*�� �Y�W �� �(��')�*� ��� O Y�W O ��W����*hV��' d \ ���)�� R�]�#�i�(') O Y�W���] %�(R�')YJ �$��k` A ���a����% �Y�� O �& %')Y*Q��*W�W$�*�b�� �+* ® � 2 B¤«!,�§v°n8�¬s© �=������'��*�^���W$�*�(h=��';Y d �� §v°Q¬s©?�?�W$�*�(`
University of Michigan Department of Mathematics Winter, 2004 Math 116 Exam 2 Problem 6 (Okefenokee) Solution
Math 116 / Exam 2 (March 20, 2017) page 3
2. [12 points] Chancelor was doodling in his coloring book one Sunday afternoon when he drewan infinity symbol, or lemniscate. The picture he drew is the polar curve r2 = 16 cos(2θ),which is shown on the axes below. (The axes are measured in inches.)
x
y
4−4
r2 = 16 cos(2θ)
a. [4 points] Chancelor decides to color the inside of the lemniscate red. Write, but do not
evaluate, an expression involving one or more integrals that gives the total area, in squareinches, that he has to fill in with red.
Solution: Notice that we are given a formula for r2 instead of r. Using symmetry, wewill calculate the area in one quarter of the lemniscate and multiply by 4. To do thiswe will integrate from θ = 0 to θ = α where α is the smallest positive number for which16 cos(2α) = 0. This gives α = π/4. Using the formula for area inside a polar curve we
see that the area is equal to 4 ·1
2
∫ π/4
0
16 cos(2θ) dθ square inches.
b. [4 points] He decides he wants to outline the right half (the portion to the right of they-axis) of the lemniscate in blue. Write, but do not evaluate, an expression involving oneor more integrals that gives the total length, in inches, of the outline he must draw inblue.
Solution: The portion of the lemniscate on the right of the y-axis corresponds to−π/4 < θ < π/4. Notice that cos(2θ) > 0 for these angles.
Implicitly differentiating r2 = 16 cos(2θ) (or directly differentiating r = 4(cos(2θ))1/2), we
find thatdr
dθ=
−4 sin(2θ)√
cos(2θ)so
(
dr
dθ
)2
=16 sin2(2θ)
cos(2θ).
Then using the arc length formula for polar coordinates we see that the length of the
blue outline will be
∫ π/4
−π/4
√
16 cos(2θ) +16 sin2(2θ)
cos(2θ)dθ inches.
c. [4 points] Chancelor draws another picture of the same lemniscate, but this time alsodraws a picture of the circle r = 2
√
2. He would like to color the area that is insidethe lemniscate but outside the circle purple. Write, but do not evaluate, an expressioninvolving one or more integrals that gives the total area, in square inches, that he mustfill in with purple.
Solution: The circle and lemniscate intersect on the right side of the y-axis when16 cos(2θ) = 8 or cos(2θ) = 1
2. This gives angles θ1 = −π/6 and θ2 = π/6.
We first find the area between the curves on the right side using the formula for the areabetween polar curves and then multiply the by 2.
The resulting total area is 2 ·1
2
∫ π/6
−π/6(16 cos(2θ)− 8) dθ square inches.
University of Michigan Department of Mathematics Winter, 2017 Math 116 Exam 2 Problem 2 (lemniscate) Solution
Math 116 / Final (April 2011) page 7
5. [14 points] Years later, after being rescued from the island, you design a machine thatwill automatically feed wood into a fire at a constant rate of 500 pounds per day. Atthe same time, as it burns, the weight of the wood pile (in pounds) decreases at a rate(in pounds/day) proportional to the current weight with constant of proportionality 1
2.
a. [3 points] LetW (t) be the weight of the wood pile t days after you start the machine.Write a differential equation satisfied by W (t).
Solution:
W′ = −0.5W + 500
b. [4 points] Find all equilibrium solutions to the differential equation in part (a).For each equilibrium solution, determine whether it is stable or unstable, and givea practical interpretation of its stability in terms of the weight of the wood pile ast → ∞.
Solution: There is one equilibrium solution, W = 1000. This equilibrium solutionis stable. This means that if your fire initially contains approximately 1000 poundsof wood, then in the long run the weight of the fire will approach 1000 pounds.
c. [7 points] Solve the differential equation from part (a), assuming that the wood pileweighs 200 pounds when you start the machine.
Solution:
dW
−0.5W + 500= dt
−2 ln | − 0.5W + 500| = t+ C
−0.5W + 500 = Ae−0.5t
W = 1000 + Ae−0.5t
W = 1000− 800e−0.5t
University of Michigan Department of Mathematics Winter, 2011 Math 116 Exam 3 Problem 5 (wood) Solution
Math 116 / Exam 2 (November 17, 2010) page 8
7. [15 points] An aquarium containing 100 liters of fresh water will be filled with a variety ofsmall fish and aquatic plants.
A water filter is installed on the tank to help remove the ammonia produced by the decomposingorganic matter generated by plants and fish in the aquarium. The filter takes water from thetank at a rate of 20 liters every hour. The water then is filtered and returned to the aquariumat the same rate of 20 liters every hour. Ninety percent of the ammonia contained in the waterthat goes through the filter is removed.
It is estimated that the fish and plants produce 30 mg of ammonia every hour. Assume theammonia mixes instantly with the water in the aquarium.
University of Michigan Department of Mathematics Fall, 2010 Math 116 Exam 2 Problem 7 (fishtank) Solution
Math 116 / Exam 2 (November 17, 2010) page 9
a. [6 points] Let Q(t) be the amount in mg of ammonia in the fish tank t hours after thefish were introduced into the aquarium. Find the differential equation satisfied by Q(t).Include its initial condition.
Solution:
dQ
dt= 30− 20
(
Q
100
)
+ (.1)20
(
Q
100
)
dQ
dt= 30− .18Q Q(0) = 0
b. [7 points] Find the amount of ammonia in the fish tank 3 hours after the fish wereintroduced into the aquarium. Make sure to include units in your answer.
Solution:
dQ
dt= 30− .18Q
dQ
30− .18Q= dt
−1
.18ln |30− .18Q| = t+ C
Q(t) =30
.18+De−.18t = 166.6 +De−.18t
using Q(0) = 0 you get Q(t) = 166.6(1− e−.18t)
Q(3) = 69.53 mg
c. [2 points] What happens to the value of Q(t) in the long run?
Solution:
limt→∞Q(t) = limt→∞ 166.6(1− e−.18t) = 166.6.
The value of Q(t) converges to 166.6 mg.
University of Michigan Department of Mathematics Fall, 2010 Math 116 Exam 2 Problem 7 (fishtank) Solution
Math 116 / Exam 2 (March 2011) page 6
4. [12 points] Another farmer notices the plague of grasshoppers has spread to his crop. Healso visits the pest control company and requests a cheaper pesticide. This new pesticide iscapable of eliminating the grasshoppers at a rate that decreases with time. Specifically, therate at which grasshoppers are killed is given by the function f(t) = 3
10(4 − t) in thousands
of grasshoppers per week at t weeks after the pesticide application. There is no pesticide re-maining after 4 weeks. Suppose there are 3000 grasshoppers at the time the pesticide is applied.
Let Q(t) the population of grasshoppers (in thousands) t weeks after this cheaper pesticide isapplied to the crop. Then for 0 ≤ t ≤ 4, Q(t) satisfies
dQ
dt=
Q
5− f(t).
a. [1 point] Is this differential equation separable?
Solution: No
b. [7 points] Using Euler’s method, fill the table with the amount of grasshoppers (in thou-sands) in the crop during the first week. Show all your computations.
t 0 1
21
Q(t) 3 2.7 2.445
Solution:
Q(0) = 3 and ∆Q = 1
2, then
Q0 = 3.
Q1 = Q0 + (Q0
5− f(0))∆Q = 2.7
Q2 = Q1 + (Q1
5− f(1
2))∆Q = 2.445
University of Michigan Department of Mathematics Winter, 2011 Math 116 Exam 2 Problem 4 (grasshoppers2) Solution
Math 116 / Exam 2 (March 2011) page 7
(problem 4 continued)
Use the slope field of the differential equation satisfied by Q(t) to answer the following ques-tions.
t
Q(t)
1 2 3 4
1
2
3
4
5
c. [2 points] Does this equation have any equilibrium solutions in the region shown? Listeach equilibrium solution and determine whether it is stable or unstable. Justify your
answer.
Solution: No equilibrium solutions. There is no y value at which all the lines have slope0.
d. [2 points] If the farmer’s goal is to kill all the grasshoppers in his crop, will the pesticidebe effective in this case? Draw the solution Q(t) on the slope field.
Solution: No
University of Michigan Department of Mathematics Winter, 2011 Math 116 Exam 2 Problem 4 (grasshoppers2) Solution
Math 116 / Exam 2 (March 25, 2019) page 11
9. [9 points] The blueprint for the Infinity Tower has been finalized, and the design of the Towerof Hanoi is accepted. Specifically:
• the tower will have infinitely many floors
• each floor has the shape of a solid cylinder of height of 3 meters
• the nth floor has radius1
2n2meters
• the ground floor corresponds to n = 1
• the tower has constant density δ kg/m3
• when construction begins, all materials are on the ground and have to be lifted to buildeach floor.
In this problem, you may assume the acceleration due to gravity is g = 9.8 m/s2.
a. [7 points] Let Wn be the work, in Joules, it takes to lift the materials to build the nthfloor and put that floor in place in the tower. Write an expression involving one or moreintegrals for each of the following.
i. W1 =
∫ 3
0π
(
1
2
)2
δgh dh
ii. W2 =
∫ 6
3π
(
1
8
)2
δgh dh =
∫ 3
0π
(
1
8
)2
δg(3 + h) dh
iii. Wn =
∫ 3n
3(n−1)π
(
1
2n2
)2
δgh dh =
∫ 3
0π
(
1
2n2
)2
δg(3(n− 1) + h) dh
b. [2 points] Write an expression involving one or more integrals and/or series that gives thetotal work it would take to build the entire tower. Your answer should not include theletter W .
Answer:
∞∑
n=1
∫ 3n
3(n−1)π
(
1
2n2
)2
δgh dh =
∞∑
n=1
∫ 3
0π
(
1
2n2
)2
δg(3(n− 1) + h) dh
University of Michigan Department of Mathematics Winter, 2019 Math 116 Exam 2 Problem 9 (Infinity Tower) Solution
Math 116 / Final (December 19, 2016) DO NOT WRITE YOUR NAME ON THIS PAGE page 11
11. [10 points] Perhaps sensing that the end is near, Steph is preparing her eventual legacy. Whileexperimenting with an accelerant called Equinate, Steph found that she could alter its heat ofcombustion by aging the Equinate in barrels. Let H(t) be the heat of combustion of Equinate,measured in hundreds of millions of Joules per kilogram, after the Equinate has been aged fort years. A graph of the derivative H
′(t) is below; note that H′(t) is linear for 1 < t < 3 and
4 < t < 5. Let R > 0 be the area of the region between the t-axis and the graph of H ′(t) for0 ≤ t ≤ 2. Let P > 0 be the corresponding area for 2 ≤ t ≤ 4.
t
H′(t)
1 3 5
-2
0
2
R
P
a. [3 points] The heat of combustion of Equinate, after aging for 5 years, is 200 million J/kg.What is the heat of combustion of Equinate that has not been aged at all? Your answermay include R and P .
Solution: The answer is 1−R+ P hundred million J/kg.
b. [3 points] Steph is storing four barrels of Equinate in the ShamCorp basement. Barrel Ahas not been aged; barrel B has been aged for 2 years; barrel C has been aged for 4 years;and barrel D has been aged for 5 years. In the spaces provided, list the barrels A, B, C,and D in increasing order of heat of combustion.
C < A < D < B
c. [4 points] At some time between 4 and 5 years of aging, the heat of combustion of Equinateis the same as if it had not been aged at all. After how many years of aging does thisoccur? Your answer may include R and P .
Solution: This occurs after 4 +√P −R years of aging.
University of Michigan Department of Mathematics Fall, 2016 Math 116 Exam 3 Problem 11 (Equinate) Solution
Math 116 / Final (April 24, 2017) do not write your name on this exam page 8
8. [8 points] A very unique sand dollar has an interesting pattern on it. The outline of the sanddollar is given by the polar equation r = 2. On the face of the sand dollar are many other
polar curves cn. For n = 1, 2, 3, . . . , the curve cn is given by the polar equation r =1
nsin(5θ).
Below is a picture of the sand dollar (left) and a zoomed in view of two of the polar curves(right).
x
y
x
y
cn
cn+1
a. [3 points] Let a0 be the area that is inside the sand dollar but outside c1.Write a formula involving one or more integrals for a0.
b. [3 points] For n ≥ 1, let an be the area inside cn but outside cn+1. Write a formulainvolving one or more integrals for an when n ≥ 1.
c. [2 points] Does the infinite series
∞X
n=0
an converge or diverge? If it converges, what is its
value? No justification is necessary.
University of Michigan Department of Mathematics Winter, 2017 Math 116 Exam 3 Problem 8 (sand dollar) Solution
page 7 of 10 Math 116, Exam 2, March 24th, 2009!
4. (50 points) The Erlang k-distribution is a probability distribution often used in mathematical modeling when events happen at a roughly (but not exactly) constant rate. It is a good model for the wait times at a telephone switchboard when calls come in on average every !
seconds. In this case, the wait time for the next k telephone calls has a probability density function that is the Erlang k-distribution
fk ,!
(x) =! k xk"1e"!x
(k "1)!x # 0
0 x < 0
$
%&
'&
.
a. For x ! 0, the Erlang k- distribution has the non-obvious cumulative distribution function,
C
k ,!(x) = 1"
e"!x (!x)n
n!n=0
k"1
# .
Using an appropriate test, show that the sum in the cumulative distribution function converges as k !" , thus verifying that
C
k ,!(x) is finite. You may assume x = 1and
! = 3. Note: saying “the distribution function must be finite, therefore it converges” will not be given credit.
As , the sum in the cumulative distribution becomes
Using the ratio test, we get since the
numerator is constant and the denominator grows to infinity. Since , the ratio test allows us to conclude that the original sum converges
University of Michigan Department of Mathematics Winter, 2009 Math 116 Exam 2 Problem 4 (switchboard) Solution
page 8 of 10 Math 116, Exam 2, March 24th, 2009!
b. A call arrives at the switchboard at 2:38:06pm. Assuming k = 1 and ! = 3 seconds, find the probability that the next phone call comes in between 2:38:08pm and 2:38:09pm.
Using the c.d.f, we simply subtract
to get 0.2355%
Using the p.d.f, we compute
which is
0.2355%
University of Michigan Department of Mathematics Winter, 2009 Math 116 Exam 2 Problem 4 (switchboard) Solution
University of Michigan Department of Mathematics Fall, 2018 Math 116 Exam 1 Problem 8 (pond) Solution
University of Michigan Department of Mathematics Fall, 2018 Math 116 Exam 3 Problem 1 Solution
Math 116 / Midterm (November 14, 2016) DO NOT WRITE YOUR NAME ON THIS PAGE page 8
9. [10 points] Suppose that f is function with the following properties:
f is differentiable f(x) > 0 for all x
∫
∞
1
f(x) dx converges.
For each of the following parts, determine whether the statement is always, sometimes, ornever true by circling the appropriate answer. No justification is needed.
a. [2 points]
∫
∞
500
1000f(x) dx converges.
ALWAYS SOMETIMES NEVER
b. [2 points]
∫
∞
1
(f(x))2/3 dx converges.
ALWAYS SOMETIMES NEVER
c. [2 points]
∫
∞
1
(f(x))3/2 dx converges.
ALWAYS SOMETIMES NEVER
d. [2 points]
∫
1
0
f(
1
x
)
dx converges.
ALWAYS SOMETIMES NEVER
e. [2 points]
∫
∞
1
f ′(x)
f(x)dx converges.
(
Note:f ′(x)
f(x)=
d
dxln(f(x)).
)
ALWAYS SOMETIMES NEVER
University of Michigan Department of Mathematics Fall, 2016 Math 116 Exam 2 Problem 9 Solution
Math 116 / Final (April 2011) page 12
10. [10 points]
a. [5 points] Determine whether the following series converge or diverge (circle youranswer). For each, justify your answer by writing what convergence rule or conver-gence test you would use to prove your answer. If you use the comparison test orlimit comparison test, also write an appropriate comparison function.
1.[2 points]
∞∑n=1
(−1)n2n
n2Converge Diverge
Solution: Diverges limn→∞(−1)n 2n
n2 6= 0.
2.[3 points]
∞∑n=1
3n− 2√n5 + n2
Converge Diverge
Solution: Converges
Limit Comparison Test bn = 1
n3
2
(or a multiple) and p series p = 32> 1.
Comparison Test bn = 3
n3
2
and p series p = 32> 1.
b. [5 points] Does the following series converge conditionally, absolutely, diverge or isit not possible to decide? Justify.
∞∑n=1
(−1)n1
n(1 + ln(n))
Solution: Is it Absolute Convergent? Using Integral test with f(x) = 1x(1+ln(x))
•f(x) ≥ 0.
•f(x) decreasing.∞∑n=1
1
n(1 + ln(n))behaves as
∫∞
1
1
x(1 + ln(x))dx
∫∞
1
1
x(1 + ln(x))dx = lim
b→∞
∫b
1
1
x(1 + ln(x))dx = lim
b→∞
ln |1 + ln(b)| = ∞
Hence∑
∞
n=11
n(1+ln(n))is not absolutely convergent.
Is it Conditionally Convergent? Since an = 1n(1+ln(n))
decreasing and converges to
zero, then by Alternating series test∑∞
n=1(−1)n 1n(1+ln(n))
converges conditionally.
University of Michigan Department of Mathematics Winter, 2011 Math 116 Exam 3 Problem 10 Solution