The radius of convergence

2 The radius of convergence

The most important statement one can make about a power series is that there exists a number, $R$ , called the radius of convergence, such that if $|x| < R$ the power series is absolutely convergent and if $|x| > R$ the power series is divergent. At the two points $x = - R$ and $x = R$ the power series may be convergent or divergent.

Key Point 11

Convergence of Power Series

For a power series $\sum_{p = 0}^{\infty} b_{p} x^{p}$ with radius of convergence $R$ then

$∙$ the series converges absolutely if $|x| < R$

$∙$ the series diverges if $|x| > R$

$∙$ the series may be convergent or divergent at $x = \pm R$

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For any particular power series $\sum_{p = 0}^{\infty} b_{p} x^{p}$ the value of $R$ can be obtained using the ratio test. We know, from the ratio test that $\sum_{p = 0}^{\infty} b_{p} x^{p}$ is absolutely convergent if

$lim_{p \to \infty} \frac{|b_{p + 1} x^{p + 1}|}{|b_{p} x^{p}|} = lim_{p \to \infty} |\frac{b_{p + 1}}{b_{p}}| |x| < 1 implying | x | < lim_{p \to \infty} |\frac{b_{p}}{b_{p + 1}}| and so R = lim_{p \to \infty} |\frac{b_{p}}{b_{p + 1}}| .$

Example 2

Find the radius of convergence of the series
$1 + \frac{x}{2} + \frac{x^{2}}{3} + \frac{x^{3}}{4} + \dots$
Investigate what happens at the end-points $x = - 1,$ $x = + 1$ of the region of
absolute convergence.

Solution

Here $1 + \frac{x}{2} + \frac{x^{2}}{3} + \frac{x^{3}}{4} + \dots = \sum_{p = 0}^{\infty} \frac{x^{p}}{p + 1}$
so

$b_{p} = \frac{1}{p + 1} ∴ b_{p + 1} = \frac{1}{p + 2}$

In this case,

$R = lim_{p \to \infty} |\frac{p + 2}{p + 1}| = 1$

so the given series is absolutely convergent if $|x| < 1$ and is divergent if $|x| > 1$ .
At $x = + 1$ the series is $1 + \frac{1}{2} + \frac{1}{3} + \dots$ which is divergent (the harmonic series). However, at $x = - 1$ the series is $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$ which is convergent (the alternating harmonic series).
Finally, therefore, the series

$1 + \frac{x}{2} + \frac{x^{2}}{3} + \frac{x^{3}}{4} + \dots$

is convergent if $- 1 \leq x < 1.$

Task!

Find the range of values of $x$ for which the following power series converges:

$1 + \frac{x}{3} + \frac{x^{2}}{3^{2}} + \frac{x^{3}}{3^{3}} + \dots$

First find the coefficient of $x^{p}$ :

$b_{p} = \frac{1}{3^{p}}$

Now find $R$ , the radius of convergence:

$R = lim_{p \to \infty} |\frac{b_{p}}{b_{p + 1}}| = lim_{p \to \infty} |\frac{3^{p + 1}}{3^{p}}| = lim_{p \to \infty} (3) = 3.$

When $x = \pm 3$ the series is clearly divergent. Hence the series is convergent only if $- 3 < x < 3.$