Calculus Exam 3 Answer Key
Instructions: Below are answers to the exam questions for use as a study resource.
Good luck on the final!
Question 1 (10 points). Compute the exact value of the following series:
∞
X 3n−1
22n−1
n=2
1
Answer: This a geometric series, so the formula we want is a · 1−r . a can be calculated by
2−1
plugging in the starting point, listed as n = 2. This gives 3 /2 2·2−1 = 3/8. Additionally,
r can be computed as r = 3/4. There are a few reasons why, but the easiest are either
computing a3 /a2 or noticing that we have 31 /22 . This gives the below answer:
3 1 31 3 3
· 3 = 1 = ·4=
8 1− 4 84 8 2
Question 2 (8 points). Compute the exact value of the following series:
X∞
cos n1 − cos n+2 1
n=1
Answer: This is a telescoping series. For telescoping, a common first step is to write out
the first few terms to look for cancelling and ultimately to compute the partial sum formula.
[cos(1)−cos( 31 )]+[cos( 12 )−cos( 14 ))]+[cos( 13 )−cos( 15 )]+. . .+[cos( n−1
1 1
)−cos( n+1 )]+[cos( n1 )−cos( n+2
1
)]
The way telescoping works, the number of positive terms at the beginning that don’t cancel
matches the number of negative terms at the end. In short, everything cancels, and we get
1 1
the partial sum formula sn = cos(1) + cos(2) − cos( n+1 ) − cos( n+2 ), for a final answer of
cos(1) + cos(2) − 2.
1
, 2
Question 3 (10 points). Determine whether the following series converges or diverges.
State what test you are using and show all work.
∞
X 1
n[ln(n)]2
n=2
While there MAY be an angle to use a comparison, the easiest by far and most expected
method is the integral test. The integral itself would use a substitution of u = ln(x) and
du = 1/x, leading to
Z ∞ Z t
1 1 1 1 1
= lim 2
du = lim − = lim − + =
2 t→∞ 2 u t→∞ u t→∞ ln(t) ln(2) ln(2)
Since the integral converges, the series will also converge by the integral test.
Question 4 (10 points). Determine whether the following series converges or diverges.
State what test you are using and show all work.
∞
X 3n2 − 2n + 4
8n4 − 2n3 + 5
n=1
The easiest method here is comparison to n12 . An inequality argument is absolutely
possible, but you could also simply do a limit comparison as below:
3n2 − 2n + 4 n2 3n4 3
lim 4 3
· = lim =
n→∞ 8n − 2n + 5 1 n→∞ 8n4 8
Since the limit is NOT equal to 0, we know that our original series converges by comparison
to n12 . Note that we’re also using the fact that n12 converges as a p-series because 2 > 1
Instructions: Below are answers to the exam questions for use as a study resource.
Good luck on the final!
Question 1 (10 points). Compute the exact value of the following series:
∞
X 3n−1
22n−1
n=2
1
Answer: This a geometric series, so the formula we want is a · 1−r . a can be calculated by
2−1
plugging in the starting point, listed as n = 2. This gives 3 /2 2·2−1 = 3/8. Additionally,
r can be computed as r = 3/4. There are a few reasons why, but the easiest are either
computing a3 /a2 or noticing that we have 31 /22 . This gives the below answer:
3 1 31 3 3
· 3 = 1 = ·4=
8 1− 4 84 8 2
Question 2 (8 points). Compute the exact value of the following series:
X∞
cos n1 − cos n+2 1
n=1
Answer: This is a telescoping series. For telescoping, a common first step is to write out
the first few terms to look for cancelling and ultimately to compute the partial sum formula.
[cos(1)−cos( 31 )]+[cos( 12 )−cos( 14 ))]+[cos( 13 )−cos( 15 )]+. . .+[cos( n−1
1 1
)−cos( n+1 )]+[cos( n1 )−cos( n+2
1
)]
The way telescoping works, the number of positive terms at the beginning that don’t cancel
matches the number of negative terms at the end. In short, everything cancels, and we get
1 1
the partial sum formula sn = cos(1) + cos(2) − cos( n+1 ) − cos( n+2 ), for a final answer of
cos(1) + cos(2) − 2.
1
, 2
Question 3 (10 points). Determine whether the following series converges or diverges.
State what test you are using and show all work.
∞
X 1
n[ln(n)]2
n=2
While there MAY be an angle to use a comparison, the easiest by far and most expected
method is the integral test. The integral itself would use a substitution of u = ln(x) and
du = 1/x, leading to
Z ∞ Z t
1 1 1 1 1
= lim 2
du = lim − = lim − + =
2 t→∞ 2 u t→∞ u t→∞ ln(t) ln(2) ln(2)
Since the integral converges, the series will also converge by the integral test.
Question 4 (10 points). Determine whether the following series converges or diverges.
State what test you are using and show all work.
∞
X 3n2 − 2n + 4
8n4 − 2n3 + 5
n=1
The easiest method here is comparison to n12 . An inequality argument is absolutely
possible, but you could also simply do a limit comparison as below:
3n2 − 2n + 4 n2 3n4 3
lim 4 3
· = lim =
n→∞ 8n − 2n + 5 1 n→∞ 8n4 8
Since the limit is NOT equal to 0, we know that our original series converges by comparison
to n12 . Note that we’re also using the fact that n12 converges as a p-series because 2 > 1
