Introduction to Power Series: Learn It 3

Giselle Miller

Introduction to Power Series: Learn It 3

Interval and Radius of Convergence

The interval of convergence is the set of all [latex]x[/latex]-values where the power series converges. When we have the third case behavior, this interval has length [latex]2R[/latex] and is centered at [latex]x = a[/latex]. The value [latex]R[/latex] represents the distance from the center to either endpoint, so we call it the radius of convergence.

Consider the familiar geometric series [latex]\displaystyle\sum_{n=0}^{\infty} x^n[/latex]. This series converges for all [latex]x[/latex] in the interval [latex](-1,1)[/latex] and diverges when [latex]|x| \geq 1[/latex]. The interval of convergence is [latex](-1,1)[/latex] , which has length 2, giving us a radius of convergence [latex]R = 1[/latex].

interval and radius of convergence

For the power series [latex]\displaystyle\sum_{n=0}^{\infty} c_n (x-a)^n[/latex]:

The interval of convergence is the set of all real numbers [latex]x[/latex] where the series converges
The radius of convergence [latex]R[/latex] is defined as:
- [latex]R = 0[/latex] if the series converges only at [latex]x = a[/latex]
- [latex]R = \infty[/latex] if the series converges for all real numbers [latex]x[/latex]
- [latex]R > 0[/latex] if the series converges for [latex]|x-a| < R[/latex] and diverges for [latex]|x-a| > R[/latex]

Figure 1 illustrates these three cases visually. In case (c), notice that the series behavior at the endpoints [latex]x = a + R[/latex] and [latex]x = a - R[/latex] requires separate investigation.

This figure has three number lines, each labeled with x. In the middle of each number line is a point labeled a. The first number line has — Figure 1. For a series [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}[/latex] graph (a) shows a radius of convergence at [latex]R=0[/latex], graph (b) shows a radius of convergence at [latex]R=\infty [/latex], and graph (c) shows a radius of convergence at R. For graph (c) we note that the series may or may not converge at the endpoints [latex]x=a+R[/latex] and [latex]x=a-R[/latex].

To determine the interval of convergence for a power series, we typically apply the ratio test.

Rules for Solving Absolute Value Inequalities

When working with convergence conditions like [latex]|x-a| < R[/latex], remember these key rules:

[latex]|x - a| \leq b[/latex] is equivalent to [latex]-b \leq x - a \leq b[/latex]
Adding or subtracting the same number to both sides preserves the inequality
Multiplying or dividing by a positive number preserves the inequality direction
Multiplying or dividing by a negative number reverses the inequality direction
If [latex]|x^n| \leq a[/latex], then [latex]-\sqrt[n]{a} \leq x \leq \sqrt[n]{a}[/latex]

Express your final answer using interval notation.

In the next example, we show the three different possibilities illustrated in Figure 1.

For each of the following series, find the interval and radius of convergence.

[latex]\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}[/latex]
[latex]\displaystyle\sum _{n=0}^{\infty }n\text{!}{x}^{n}[/latex]
[latex]\displaystyle\sum _{n=0}^{\infty }\frac{{\left(x - 2\right)}^{n}}{\left(n+1\right){3}^{n}}[/latex]

To check for convergence, apply the ratio test. We have

[latex]\begin{array}{cc}\hfill \rho & =\underset{n\to \infty }{\text{lim}}|\frac{\frac{{x}^{n+1}}{\left(n+1\right)\text{!}}}{\frac{{x}^{n}}{n\text{!}}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{x}^{n+1}}{\left(n+1\right)\text{!}}\cdot \frac{n\text{!}}{{x}^{n}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{x}^{n+1}}{\left(n+1\right)\cdot n\text{!}}\cdot \frac{n\text{!}}{{x}^{n}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{x}{n+1}|\hfill \\ & =|x|\underset{n\to \infty }{\text{lim}}\frac{1}{n+1}\hfill \\ & =0<1\hfill \end{array}[/latex]

for all values of x. Therefore, the series converges for all real numbers x. The interval of convergence is [latex]\left(\text{-}\infty ,\infty \right)[/latex] and the radius of convergence is [latex]R=\infty[/latex].
Apply the ratio test. For [latex]x\ne 0[/latex], we see that

[latex]\begin{array}{cc}\hfill \rho & \hfill =\underset{n\to \infty }{\text{lim}}|\frac{\left(n+1\right)\text{!}{x}^{n+1}}{n\text{!}{x}^{n}}|\\ & =\underset{n\to \infty }{\text{lim}}|\left(n+1\right)x|\hfill \\ & =|x|\underset{n\to \infty }{\text{lim}}\left(n+1\right)\hfill \\ & =\infty .\hfill \end{array}[/latex]

Therefore, the series diverges for all [latex]x\ne 0[/latex]. Since the series is centered at [latex]x=0[/latex], it must converge there, so the series converges only for [latex]x\ne 0[/latex]. The interval of convergence is the single value [latex]x=0[/latex] and the radius of convergence is [latex]R=0[/latex].
In order to apply the ratio test, consider

[latex]\begin{array}{cc}\hfill \rho & =\underset{n\to \infty }{\text{lim}}|\frac{\frac{{\left(x - 2\right)}^{n+1}}{\left(n+2\right){3}^{n+1}}}{\frac{{\left(x - 2\right)}^{n}}{\left(n+1\right){3}^{n}}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{\left(x - 2\right)}^{n+1}}{\left(n+2\right){3}^{n+1}}\cdot \frac{\left(n+1\right){3}^{n}}{{\left(x - 2\right)}^{n}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{\left(x - 2\right)\left(n+1\right)}{3\left(n+2\right)}|\hfill \\ & =\frac{|x - 2|}{3}.\hfill \end{array}[/latex]

The ratio [latex]\rho <1[/latex] if [latex]|x - 2|<3[/latex]. Since [latex]|x - 2|<3[/latex] implies that [latex]-31[/latex] if [latex]|x - 2|>3[/latex]. Therefore, the series diverges if [latex]x<-1[/latex] or [latex]x>5[/latex]. The ratio test is inconclusive if [latex]\rho =1[/latex]. The ratio [latex]\rho =1[/latex] if and only if [latex]x=-1[/latex] or [latex]x=5[/latex]. We need to test these values of x separately. For [latex]x=-1[/latex], the series is given by

[latex]\displaystyle\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}}{n+1}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots[/latex].

Since this is the alternating harmonic series, it converges. Thus, the series converges at [latex]x=-1[/latex]. For [latex]x=5[/latex], the series is given by

[latex]\displaystyle\sum _{n=0}^{\infty }\frac{1}{n+1}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots[/latex].

This is the harmonic series, which is divergent. Therefore, the power series diverges at [latex]x=5[/latex]. We conclude that the interval of convergence is [latex]\left[-1,5\right)[/latex] and the radius of convergence is [latex]R=3[/latex].