Radius and Interval of Convergence

A Taylor polynomial has finitely many terms. Letting the terms go on forever gives a series whose terms involve

x

, and whether it converges can depend on the value of

x

Power series

In the middle case the series converges on an open interval from

r - R

r + R

, and the two endpoints have to be checked separately. The interval of convergence is that open interval plus whichever endpoints the series converges at.

The ratio test usually finds

R

. Apply it to the absolute values of the terms, with

x

treated as a fixed number. When the limit

L

exists, it depends on

x

, and the series converges where

L < 1

and diverges where

L > 1

, which fixes

R

. At the endpoints

L = 1

, so the ratio test can't decide them, and another test has to.

Example 1

Find the radius and the interval of convergence of

\sum_{n=1}^{\infty} \frac{n\,(x - 5)^n}{4^n (n^2 + 1)}

Call the

n

th term

a_n

. For

x \ne 5

, take the ratio of consecutive terms, in absolute value.

\begin{aligned} \left|\frac{a_{n+1}}{a_n}\right| &= \frac{n + 1}{n}\cdot\frac{n^2 + 1}{(n + 1)^2 + 1} \\[4pt] &\qquad \cdot\frac{|x - 5|}{4} \\[4pt] &\to 1 \cdot 1 \cdot \frac{|x - 5|}{4} = \frac{|x - 5|}{4} \end{aligned}

The series converges when

|x - 5| < 4

and diverges when

|x - 5| > 4

. The center is 5, the radius is

R = 4

, and the open interval is

1 < x < 9

x = 9

the series is

\sum \frac{n}{n^2 + 1}

. Its terms behave like

\frac{1}{n}

, and

\lim_{n\to\infty} \frac{n/(n^2 + 1)}{1/n} = 1

. The harmonic series diverges, so this one diverges by limit comparison.

x = 1

the series is

\sum \frac{(-1)^n n}{n^2 + 1}

. The terms alternate in sign. The sizes

\frac{n}{n^2 + 1}

approach 0, and they decrease, since the derivative of

\frac{x}{x^2 + 1}

\frac{1 - x^2}{(x^2 + 1)^2} \le 0

for

x \ge 1

. It converges by the alternating series test.

The interval of convergence is

1 \le x < 9

Radius 0 and radius ∞

Example 2

Find the radius of convergence of each series.

(a)

\displaystyle\sum_{n=0}^{\infty} \frac{3^n x^{2n+1}}{(2n + 1)!}

(b)

\displaystyle\sum_{n=1}^{\infty} n^n x^n

(a) Only odd powers of

x

appear, so take the ratio of whole terms,

x

included. At

x = 0

every term is 0. For any fixed

x \ne 0

\begin{aligned} &\left|\frac{3^{n+1} x^{2n+3}}{(2n + 3)!}\cdot\frac{(2n + 1)!}{3^n x^{2n+1}}\right| \\[4pt] &\qquad = \frac{3x^2}{(2n + 3)(2n + 2)} \to 0 \end{aligned}

The limit is

0 < 1

for every

x

, so the series converges everywhere and

R = \infty

(b) For any fixed

x \ne 0

\begin{aligned} &\left|\frac{(n + 1)^{n+1} x^{n+1}}{n^n x^n}\right| \\[4pt] &\qquad = (n + 1)\Big(\frac{n + 1}{n}\Big)^n |x| \to \infty \end{aligned}

since

\big(\frac{n + 1}{n}\big)^n \to e

. The series diverges for every

x \ne 0

, so

R = 0

. It converges only at its center,

x = 0

Differentiating and integrating

On its open interval of convergence, a power series with

R > 0

defines a function

f

. That series can be differentiated or integrated term by term, giving a series for

f'

or for an antiderivative of

f

. The new series has the same radius of convergence, but its behavior at the endpoints can change, so they have to be checked again.

Example 3

Let

f(x) = \sum_{n=1}^{\infty} \frac{n\,(x - 5)^n}{4^n (n^2 + 1)}

, the series from Example 1. Find the interval of convergence of the series for

f'(x)

Differentiate term by term.

f'(x) = \sum_{n=1}^{\infty} \frac{n^2 (x - 5)^{n-1}}{4^n (n^2 + 1)}

The radius is still 4, so check

x = 1

and

x = 9

again. At both,

|x - 5| = 4

, so the terms have size

\frac{n^2 \cdot 4^{n-1}}{4^n (n^2 + 1)} = \frac{n^2}{4(n^2 + 1)}

, which approaches

\frac{1}{4} \ne 0

. Both endpoints diverge by the nth term test.

	$f(x)$	$f'(x)$
At $x = 1$	converges	diverges
At $x = 9$	diverges	diverges
Interval	$[1, 9)$	$(1, 9)$