NAMES:
MATH 152
April 8, 2015
QUIZ 8
• Show all your work and indicate your final answer clearly. You will be graded not merely
on the final answer, but also on the work leading up to it.
1. (3 points) Test the series for convergence or divergence.
∞
X
(−1)n−1
n=1
ln(n + 4)
Solution: Since
1
= 0,
ln(n + 4)
the alternating series test gives that the series converges.
lim
n→∞
2. (3 points) Determine whether the series is absolutely convergent, conditionally convergent
or divergent.
∞
X
(−2)n n8
n!
n=1
Solution: To show absolute convergence, we use the ratio test:
an+1 (−2)n+1 (n + 1)8 n! lim
= lim n→∞ an n→∞
(−2)n n8 (n + 1)! (−2)(n + 1)8 = lim 8
n→∞
n (n + 1) =0
(since numerator is degree 8 and denominator is degree 9)
and so the series is absolutely convergent by the ratio test.
NAMES:
MATH 152
April 8, 2015
3. (3 points) Determine whether the series is absolutely convergent, conditionally convergent,
or divergent
∞
X
n
(−1)n √
3
n +4
n=1
Solution: First note that the series converges by the alternating series test since
n
lim √
=0
3
n→∞
n +4
n
and √n+4
is strictly decreasing.
To check absolte convergence, we use the definition of absolute convergence and check if
∞ ∞
X
X
n
n
(−1)n √
=
√
3
3
n +4
n +4
n=1
n=1
P∞ 1
converges. Using the limit comparison with n=1 √n gives
√
√ n
n n
n3/2
n3 +4
√
√
lim
=
lim
=
lim
= 1 (L’Hospital’s Rule)
n→∞ √1
n→∞
n3 + 4 n→∞ n3 + 4
n
P
P
P∞
√1 and
√ n
√1
both converge or diverge. Since ∞
gives that ∞
n=1 n diverges by
n=1 nP
n=1 n3 +4P
∞
∞
the p-test, so does n=1 √nn3 +4 and n=1 (−1)n √nn3 +4 is not absolutely convergent.
Thus the series is conditionally convergent.