Analysis Problems #8
Solutions
1. Test each of the following series for convergence:
∞
∑
e1/n
n=1
n2
∞
∑
e1/n
,
n=1
n
∞
∑
n+1
,
n=1
2n
.
• When it comes to the first series, we use the comparison test. Since
∞
∑
e1/n
n=1
n2
∞
∑
e
≤
,
n2
n=1
the given series is smaller than a convergent p-series, so it converges as well.
• To test the second series for convergence, we use the limit comparison test with
an =
e1/n
,
n
bn =
1
.
n
Note that the limit comparison test is, in fact, applicable here because
an
lim
= lim e1/n = e0 = 1.
n→∞ bn
n→∞
∑∞
∑
Since n=1 bn is a divergent p-series, the series ∞
n=1 an must also diverge.
• To test the third series for convergence, we use the ratio test. In this case,
an+1
n + 2 2n
1
= lim
· n+1 =
n→∞ an
n→∞ n + 1 2
2
L = lim
is strictly less than 1, so the third series converges by the ratio test.
2. Test each of the following series for convergence:
(
)n
∞
∞
∑
∑
1
1
n!
· 1+
,
.
n
n
n
n
n=1
n=1
• When it comes to the first series, we use the limit comparison test with
(
)n
1
e
1
,
bn = .
an = · 1 +
n
n
n
Note that the limit comparison test does apply here because
(
)n
an
1
1
lim
= lim · 1 +
= 1.
n→∞ bn
n→∞ e
n
∑∞
∑
Since ∞
n=1 an must also diverge.
n=1 bn is a divergent p-series, the series
• To test the second series for convergence, we use the ratio test. In our case,
(
)n
(n + 1)!
nn
n
an+1
nn
=
·
=
=
an
n!
(n + 1)n+1
(n + 1)n
n+1
and this implies that
an+1
L = lim
= lim
n→∞ an
n→∞
(
n
n+1
)n
=
1
.
e
Since e > 1, we have L < 1, so the given series converges by the ratio test.
3. Find the radius of convergence for each of the following power series:
∞
∑
nx2n
n=0
4n
,
∞
∑
n! xn
n=0
(2n)!
.
• When it comes to the first power series, we have
an+1 n + 1 x2n+2 4n
n+1 2
an = n · x2n · 4n+1 = 4n · x
and so
2
an+1 = lim n + 1 · x2 = x .
L = lim n→∞
an n→∞ 4n
4
In particular, the given series converges when x2 < 4 and it diverges when x2 > 4, so
its radius of convergence is R = 2.
• When it comes to the second power series, we have
an+1 (n + 1)! |x|n+1
(2n)!
·
·
an =
n!
|x|n (2n + 2)!
and so
an+1 (n + 1) · |x|
= lim
L = lim = 0.
n→∞
n→∞ (2n + 1)(2n + 2)
an
Thus, the series converges for all x ∈ R and its radius of convergence is R = ∞.
4. Assuming that |x| < 1, show that the power series
f (x) =
∞
∑
(−1)n−1 xn
n=1
n
converges and that f (x) = log(1 + x). Hint: differentiate.
• First of all, we use the ratio test to find that
n+1
an+1 n
= lim |x|
L = lim ·
= |x|.
n
n→∞
n→∞
an
|x|
n+1
This shows that f (x) converges whenever |x| < 1 and it also implies that
′
f (x) =
∞
∑
(−1)n−1 nxn−1
n=1
n
=
∞
∑
n−1
(−x)
n=1
=
∞
∑
(−x)n .
n=0
Using the formula for a geometric series, we now get
f ′ (x) =
1
1
=
1 − (−x)
1+x
=⇒
f (x) = log(1 + x) + C.
Since f (0) = 0, however, we must have 0 = log 1 + C = C, so f (x) = log(1 + x).