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solutions to some of the exercises in section 11.
- Find the radius of convergence and the interval of convergence of the series
n=
(−1)n^ xn n+
Solution. If we set an =
(−1)nxn n+
then we have
an+
an
n+ x
n+
n + 2
n + 1
(−1)nxn
= |x|
n + 1
n + 2
→ |x| as n → ∞.
Thus by the ratio test we conclude that
n=
(−1)nxn n+
is convergent when |x| < 1 and
divergent when |x| > 1. Therefore the radius of convergence is 1. To find the interval
of convergence we need to check the endpoints (x = ±1) separately:
x = 1 When x = 1 the power series becomes
n=
(−1)n n+
. Set bn =
1 n+
. Since
- bn ≥ 0 for all n ≥ 1.
- bn =
1 n+
1 n+
for n ≥ 1. Thus {bn} is a decreasing sequence.
bn = lim n→∞
1 n+
Thus by the alternating series test we conclude the series
n=
(−1)n n+
is convergent.
x = − 1 When x = −1 the power series becomes
n=
1 n+
. Set bn =
1 n
. Notice that bn
and
1 n+
are both positive for all n > 0. Also,
1 n+
bn
n
n + 1
→ 1 as n → ∞
and the series
n=
1 n
is divergent (as it’s the harmonic series). Thus by the limit
comparison test we conclude that
n=
1 n+
is divergent, hence
n=
1 n+
is divergent.
Thus the interval of convergence is (− 1 , 1].
- Find the radius of convergence and the interval of convergence of the series
n=
nx
n .
Solution. If we set an =
nx
n then we have
an+
an
n + 1x
n+
√ nx
n
= |x|
n + 1
n
= |x|
n
→ |x| as n → ∞.
Thus by the ratio test we conclude that
n=
(−1)nxn n+
is convergent when |x| < 1 and
divergent when |x| > 1. Therefore the radius of convergence is 1. To find the interval
of convergence we need to check the endpoints (x = ±1) separately:
x = 1 When x = 1 the power series becomes
n=
n. Since lim n→∞
n = ∞ 6 = 0 by the
test for divergence we conclude the series
n=
n is divergent.
x = − 1 When x = −1 the power series becomes
n=
n
n. Since lim n→∞
n
n 6 = 0
by the test for divergence we conclude the series
n=
n
n is divergent.
Thus the interval of convergence is (− 1 , 1).
- Find the radius of convergence and the interval of convergence of the series
n=
xn 5 nn^5
Solution. If we set an =
xn 5 nn^5
then we have
an+
an
x
n+
n+ (n + 1)
5
n n
5
x
n
|x|
n
n + 1
|x|
1 + 1/n
|x|
as n → ∞.
Thus by the ratio test we conclude that
n=
xn 5 nn^5
is convergent when |x| < 5 and
divergent when |x| > 5. Therefore the radius of convergence is 5. To find the interval
of convergence we need to check the endpoints (x = ±5) separately:
x = 5 When x = 5 the power series becomes
n=
1 n^5
which is a p-series with p = 5 > 1
and hence convergent.
x = − 5 When x = −5 the power series becomes
n=
(−1)n n^5
. Since
n=
(−1)n n^5
∣ =^
n=
1 n^5
is convergent we see that
n=
(−1)n n^5
is absolutely convergent and thus convergent.
Thus the interval of convergence is [− 5 , 5].
- Find the radius of convergence and the interval of convergence of the series
n=
n xn 4 n^ ln(n)
Solution. If we set an = (−1)
n xn 4 n^ ln(n)
then we have
lim n→∞
an+
an
= lim n→∞
x
n+
n+ ln(n + 1)
n ln(n)
x
n
|x|
lim n→∞
ln(n)
ln(n + 1)
L’Hopital ↓ =
|x|
lim n→∞
1 n
1 n+
|x|
lim n→∞
n + 1
n
|x|
Thus by the ratio test we know the power series
n=
n x
n
4 n^ ln(n)
is convergent if
|x| < 4 and divergent if |x| > 4. In other words the radius of convergence is 4. To find
the interval of convergence we need to check the endpoints (x = ±4) separately:
x = 4 When x = 4 the power series becomes
n=
n 4 n 4 n^ ln(n)
n=
n 1 ln(n)
. Set
bn =
1 ln(n)
. Then we have
x =
5 2
When x = 3 +
1 2
the power series becomes
n=
(−2)n √ n
1 2
n
n=
√^1 n
which is a
p-series with p =
1 2
< 1 and hence divergent.
Thus the interval of convergence is
5 2
7 2
]
