Day 11-2: Convergence of Power Series

1 Power Series
11-2: Convergence of Power Series
Prakash Balachandran
Department of Mathematics
Duke University
March 23, 2010
• Definition: A power series about x = a is a sum of constants times powers of (x − a) :
C0 + C1(x − a) + C2(x − a) 2 + · · · + Cn(x − a) n + · · · =
• In the definition of a power series, a is a constant.
∞
Cn(x − a) n .
• For any fixed value of x, the power series ∞
n=0 Cn(x − 1) n is a series of numbers as we’ve considered
before. To investigate the convergence of a power series, we consider the partial sums Sn(x) = C0 +
C1(x − 1) + · · · + Cn(x − a) n and ask whether for each fixed x, {Sn(x)} tends to some limit as n → ∞.
• Recall the geometric series:
n=0
∞
x n . (1)
n=0
This is a power series about x = 0 where Cj = 1 for every j.
– Setting Sn(x) = n j=0 xj , we find that Sn(x) → 1 if −1 < x < 1.
1−x
– What if x = −1, 1?
∗ If we plug in x = 1, then {Sn(1)} = {n + 1} which obviously increases without bound, and so
the series diverges.
∗ If we plug in x = −1 then {Sn(−1)} oscillates between 0 and 1 and so diverges.
– Notice the convergence of this power series to a finite number is non-trivial: it converges to 1
1−x only
for −1 < x < 1. Can we develop theorems or tests that tell us when a general power series converges?
• Example:
– Determine whether the power series ∞ x
n=0
n
2n converges or diverges for
1. x = −1.
∗ Plugging in x = −1, we find ∞ the series converges to
1
1−(− 1
2)
n=0
= 2
3 .
1
− 1
2
n 1
which is a geometric series with ratio − . Thus, 2

2. x = 3.
2 Convergence
∗ Plugging in x = 3, we find ∞ n=0
Thus, the series diverges at x = 3.
2.1 The Interval of Convergence
3 n
which is a geometric series with ratio bigger than 1.
2
• In the previous section, we discussed convergence and divergence of power series. We noticed that, at least
in the case of the geometric series, there was an interval in which it converged, but it didn’t converge at the
endpoints. Is this a particular phenomenon of the geometric series, or does it hold in more generality?
• Definitions:
– If a power series converges only for x = a, then the radius of convergence is defined to be R = 0.
– If the power series converges for all values of x, then the radius of convergence is defined to be
R = ∞.
– If the power series converges for values of x for which |x − a| < R, or a − R < x < a + R, then the
radius of convergence is defined to be R.
– The interval of convergence is the interval (a − R, a + R) including and endpoint where the power
series converges.
• In the case of the geometric series, ∞
n=0 xn , the radius of convergence is 1, and the interval of convergence
is (−1, 1).
• As promised, we have a theorem that computes convergence over intervals::
Theorem 1 (Method for Computing the Radius of Convergence) To compute the radius of convergence, R,
for the power series ∞
n=0 Cn(x − a) n , use the ratio test with an = Cn(x − a) n :
1. If limn→∞ |an+1|
|an|
2. If limn→∞ |an+1|
|an|
3. If limn→∞ |an+1|
|an|
= ∞ then R = 0.
= 0 then R = ∞.
= K|x − a| where K is finite and nonzero, then R = 1
K .
4. If the limn→∞ |an+1|
|an| fails to exist, then we don’t know anything.
Proof:
∞
Cn(x − a) n
≤
and we have that
|an+1|
lim
n→∞ |an|
n=0
∞
|Cn||x − a| n =
n=0
|Cn+1||x − a|
= lim
n→∞ |Cn|
2
∞
|an|
n=0
|Cn+1|
= |x − a| lim
n→∞ |Cn|
= L|x − a|.

So, depending on the value of L, the ratio test gives us when the power series converges.
• Examples:
1. Show that the power series
converges for all values of x.
– Cn = 1
n!
all x.
⇒ limn→∞ |an+1|
|an|
1 + x + x2
2!
= |x| limn→∞ 1
n+1
+ · · ·
2. Determine the radius and interval of convergence of the series
– Cn = (−1)n−1
n
(x − 1) −
(x − 1)2
2
+ (x − 1)3
3
|an+1|
⇒ limn→∞ |an|
is R = 1, and the interval of convergence is (0, 2).
3. Find the interval of convergence of the series
n−1 x2n−1
– an = (−1) (2n−1)!
every value of x.
x − x3
3!
+ x5
5!
= |x − 1| limn→∞ n
n+1
− x7
7!
⇒ |an+1|
|an| = x 2
(2n+1)2n
= 0, so that R = ∞. Thus, the series converges for
n−1 (x − 1)n
+ · · · + (−1) + · · ·
n
= |x − 1|. Thus, the radius of convergence
x2n−1
+ · · · + (−1)n−1 + · · ·
(2n − 1)!
→ 0 as n → ∞. Thus, the power series converges for
• In the last example, the moral is: if some of the Cn are zero, rewrite the series and pick one in which the
terms are nonzero.
2.2 Endpoints of the Interval
• The ratio test doesn’t tell us what happens to a power series at the endpoints of its interval of convergence.
Often, we can just plug in x = a ± R and convert the power series to a series of numbers, and use the tests
we already know to establish convergence/divergence at these endpoints.
• Example: Look at the previous example (2). Does it converge at the end points?
– Plugging in x = 2, we obtain the series 1 − 1 1 1 + − 2 3 4
series, and so converges by the alternating series test.
– Plugging in x = 0, we obtain the series −1 − 1
2
series, which diverges.
3
+ · · · + (−1)n−1
n
+ · · · which is an alternating
1 1
1
− − − · · · − − · · · which is a negative harmonic
3 4 n