Math 231 Worksheet #13 Solutions
P
P∞
1
1
1. (a) Use the comparison test with ∞
n=1 n2 to show
n=1 n2 +1 converges.
Solution:
1
≥ n21+1 for all n ≥ 1
n2
P∞ 1
We know that
n=1 n2 converges by the integral test, so the original series
P
∞
1
n=1 n2 +1 must also converge.
P
P∞
1
1
(b) Why can’t you use the comparison test with ∞
n=2 n2 to show
n=2 n2 −1 converges?
Solution:
P∞ 1
Because n12 < n21−1 for all n ≥ 2, the convergence of the series
n=1 n2 says
P∞
1
nothing about the convergence of the original series n=1 n2 −1 .
P∞
P
1
1
(c) Use the limit comparison test with ∞
n=2 n2 −1 converges.
n=2 n2 to show
Solution:
1
2
an = n12 , bn = n21−1 , limn→∞ n12 = limn→∞ nn−1
=1
2
n2 −1
P
1
By that result, both series converge because we know that ∞
n=1 n2 converges.
P
P∞ 1
1
2. (a) Use the comparison test with ∞
n=2 n to show
n=2 n−1 diverges.
Solution:
1
1
< n−1
for all n ≥ 2
n
P∞ 1
P
1
diverges
by
the
integral
test,
so
the
original
series
We know that ∞
n=2 n−1
n=1 n
must also diverge.
P
P∞ 1
1
(b) Why can’t you use the comparison test with ∞
n=1 n to show
n=1 n+1 diverges?
Solution:
P
1
1
for all n ≥ 1, divergence of the series ∞
Because n1 > n+1
n=1 n says nothing about
P∞ 1
the divergence of the original series n=1 n+1 .
P
P∞ 1
1
(c) Use the limit comparison test with ∞
n=1 n to show
n=1 n+1 diverges.
Solution:
1
1
, limn→∞ n1 = limn→∞ n+1
=1
an = n1 , bn = n+1
n
n+1
P
1
By that result, both series diverge because we know that ∞
n=1 n diverges.
3. Determine whether the series converges or diverges.
∞
X
n3
(a)
n4 − 1
n=2
Solution:
3
n3
> nn4 = n1
n4 −1
P∞ 1
P∞
n=2 n diverges by the integral test, so the original series
n=2
n3
n4 −1
also diverges.
(b)
(c)
∞
X
n−1
√
2 n
n
n=1
Solution:
n
1
n−1
√ = n−1
< n5/2
= n3/2
n2 n
n5/2
P∞ 1
P∞
n=1 n3/2 converges by the integral test, so the original series
n=1
converges.
∞
X
1 + sin n
Solution:
2
1 n
1+sin n
<
=
2
·
n
n
10
10
10
P
1 n
converges,
geometric series with r =
2P· ∞
n=0 10
∞ 1+sin n
also converges.
n=0 10n
1
10
< 1, so the original series
2
3
< 1, so the original series
∞
X
n + 4n
n=0
n + 6n
Solution:
n+4n
n+4n
4n +4n
4n
2 n
≤
≤
=
2
·
=
2
·
n
n
n
n
n+6
6
6
6
3
P∞ 2 n
2P· n=0 3 converges, geometric series with r =
∞ n+4n
n=0 n+6n also converges.
(e)
also
10n
n=0
(d)
n−1
√
n2 n
∞
X
e1/n
n=1
n
Solution:
e1/n
> n1
n
P∞ 1
P∞
n=1 n diverges by the integral test, so the original series
n=1
e1/n
n
also diverges.
∞
X
n!
(f)
nn
n=1
Solution:
n!
= n·(n−1)·(n−2)·...·2·1
≤ n2 · n1 = n22
nn
P∞ n·n·n·n·...·n·n
P∞
2 · n=1 n12 converges by the integral test, so the original series
n=1
converges.
n!
nn
also