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MATH 31B DISCUSSION: INTERVAL OF CONVERGENCE OF
POWER SERIES
General procedure:
- Write down the center c of the power series.
- Find the radius of convergence R by using Ratio Test.
- Check convergence at the end points c + R, c โ R, using any test possible.
Example 0.1. Determine the interval of convergence:
โ^ โ
n=
n
p x
n
where p > 0.
Solution: This is a power series centered at c = 0. First, we find the radius of
convergence R using Ratio Test:
ฯ = lim nโโ
|(n + 1)
p x
n+ |
|npxn|
= lim nโโ
(n + 1)
p
np^
|x|
= lim nโโ
n + 1
n
)p
|x|
= lim nโโ
n
)p
|x|
= |x|
By the Ratio Test, the series converges if ฯ = |x| < 1 and diverges if ฯ = |x| > 1.
This means the radius of convergence R = 1. It remains to determine whether the
series converges at x = 1, โ1.
When x = 1, the series equals
โโ
n=
n
p
which diverges by Divergence Test, since n
p does not got to 0 as n โ โ. When
x = โ1, the series equals
โโ
n=
n
p (โ1)
n
which also diverges by Divergence Test, since n
p (โ1)
n does not got to 0 as n โ โ.
Thus, the interval of convergence is the open interval (โ 1 , 1).
1
Example 0.2. Determine the interval of convergence:
โ^ โ
n=
(โ2)n(x โ 2)n
n^2
Solution: This is a power series centered at c = 2. First, we find the radius of
convergence R using Ratio Test:
ฯ = lim nโโ
(โ2)n+1(xโ2)n+ (n+1)^2
(โ2)n(xโ2)n n^2
= lim nโโ
n
2
(n + 1)^2
ยท 2 |x โ 2 |
= lim nโโ
n
n + 1
ยท 2 |x โ 2 |
= lim nโโ
1 n
ยท 2 |x โ 2 |
= 2|x โ 2 |
By the Ratio Test, the series converges if ฯ = 2|x โ 2 | < 1 and diverges if ฯ =
2 |x โ 2 | > 1. Solving these inequalities, the series converges if |x โ 2 | <
1 2 and
diverges if |x โ 2 | >
1 2
. (This means the radius of convergence R =
1 2
.) It remains to
determine whether the series converges when |x โ 2 | =
1 2
When x โ 2 =
1 2 , the series equals
โ^ โ
n=
n
n^2
which converges, either by the Alternating Series Test or by the p-test for its absolute
value series. When x โ 2 = โ
1 2 ,
โ^ โ
n=
n^2
which converges by the p-test.
Thus, the interval of convergence is the closed interval [2 โ
1 2
1 2
] = [
3 2
5 2
].
Example 0.3. Determine the interval of convergence:
โ^ โ
n=
50 x
3 n+
2 n^
In the last step we calculated the limits
lim nโโ
n โ n + 1
= lim nโโ
1 n
lim nโโ
ln(n)
ln(n + 1)
= lim nโโ
ln(n)
ln(n) + ln
n+ n
) = lim nโโ
1 ln(n) ln(1 +^
1 n )
You could also calculate these limits by Lโhospitalโs Rule.
By the Ratio Test, the series converges if ฯ = |x| < 1 and diverges if ฯ = |x| > 1.
(This means the radius of convergence R = 1.) It remains to determine whether the
series converges when |x| = 1.
When x = 1, the series equals
โ^ โ
n=
n ln(n)
which diverges by Direct Comparison Test:
n ln(n)
n
for n large enough, and
the series
โ^ โ
n=
n
diverges by the p-test.
When x = โ1, the series equals
โ^ โ
n=
n
โ n ln(n)
which converges by Alternating Series Test.
Thus, the interval of convergence is the (half-open) interval [โ 1 , 1).
Example 0.5. Determine the interval of convergence:
โ โ
n=
x
n^2
n!
Solution: This is a power series centered at c = 0. First, we find the radius of
convergence R using Ratio Test:
ฯ = lim nโโ
xn+
2
(n+1)!
xn
2
n!
= lim nโโ
|x|
(n+1)^2 โn^2
n + 1
= lim nโโ
|x|
2 n+
n + 1
โ if |x| > 1
0 if |x| โค 1
By the Ratio Test, the series converges if ฯ < 1 and diverges if ฯ > 1, which
corresponds to |x| โค 1 and |x| > 1 respectively. Thus, the radius of convergence
is R = 1, and the interval of convergence is [โ 1 , 1]. (Notice that in this problem,
we donโt need to separately consider the end points x = โ 1 , 1 because we already
conclude from the Ratio Test that the series converges at these end points. )
