Calculus II Sample Exam 3

```Calculus II
Sample Exam 3: Chapter 10
1. Does the sequence described by an = (−1)n
n2
converge?
2n2 − 5
n2
1
n2
n
=
,
the
terms
(−1)
do not coverge.
n→∞ 2n2 − 5
2
2n2 − 5
∞
X
(−2)n
2. To what number does
converge?
7n+1
n=1
Answer. This is a geometric series missing the first term:
n
n
∞
∞
∞
X
X
(−2)n
1 −2
1 X 1 −2
1 1
1
1 1
2
=− + =−
=
=− +
=− +
n+1
7
7
7
7 n=0 7
7
7 7 1 − (−2/7)
7 9
63
n=1
n=1
3. Does
∞
X
1
converge?
n(ln n)3
n=2
Z
2
∞
1
dx = lim
t→∞
x(ln x)3
Z
ln t
u−3 du =
ln 2
1
,
2(ln 2)2
so the series converges.
4. Does
∞
X
n2 − 2
converge?
n3 + n + 2
n=2
Answer. This “looks like” the harmonic series, since n2 /n3 = 1/n, so we guess that it
diverges. We’d like to see that the terms are larger than terms that look like 1/n. It is
true that for large enough n:
n2 − 2
n2 /2
1
≥
=
,
3
3
n +n+2
3n
6n
because eventually n2 − 2 &gt; n2 /2 and n3 + n + 2 &lt; 3n3 . So the series diverges because
P
1/(6n) diverges.
Calculus II—Sample Exam 3: Chapter 10
5. Does
2
∞
X
(−1)n
converge absolutely, converge conditionally, or diverge?
n ln n
n=2
Answer. It converges conditionally. It converges by the alternating series test, but it
∞
X
1
doesn’t converge absolutely because
diverges by the integral test.
n ln n
n=2
6. Does
∞
X
n2
converge?
n!
n=1
Answer. Yes, by the ratio test:
(n + 1)2 n!
lim
= lim
n→∞ (n + 1)! n2
n→∞
n+1
n
2
1
= 0.
n+1
7. Find the interval of convergence and radius of convergence for
∞
X
3n n
x .
n
n=1
3n+1 n
|x| = 3|x|,
n→∞ n + 1 3n
lim
so the radius of convergence is 1/3. When x = 1/3 the series is the harmonic series, so
it diverges; when x = −1/3 the series is the alternating harmonic series, so it converges;
thus the interval of convergence is [−1/3, 1/3). On the real exam, as usual, (−1/3, 1/3)
would be acceptable.
∞
X
n3
8. Find the interval of convergence and radius of convergence for
(x − 2)n .
n
2
n=0
(n + 1)3 2n
lim
|x − 2| = lim
n→∞ 2n+1
n→∞
n3
n+1
n
3
1
1
|x − 2| = |x − 2|,
2
2
so the radius of convergence is 2 and the interval of convergence is (0, 4). It is not hard
to see that it does not converge at the endpoints 0 and 4.
Calculus II—Sample Exam 3: Chapter 10
3
9. Find a power series representation for x2 ex ; find the radius of convergence for your series.
x2
∞
∞
X
X
xn
xn+2
=
.
n!
n!
n=0
n=0
Using the ratio test:
n!
1
|x| =
|x| = 0,
n→∞ (n + 1)!
n+1
lim
so the radius of convergence is ∞.
10. Find a power series representation for cos(x3 ); find the radius of convergence for your series.
Answer. It’s difficult to do this by computing derivatives of cos(x3 ), but easy to do by
starting with the series for the cosine:
∞
X
(−1)n 2n
cos x =
x ,
(2n)!
n=0
and substituting x3 for x:
cos x3 =
∞
X
(−1)n 6n
x .
(2n)!
n=0
Since the series for cosine converges for all numbers, so does the second series, so the