Form and Convergence of a Power Series

Learning Outcomes

Identify a power series and provide examples of them
Determine the radius of convergence and interval of convergence of a power series

Form of a Power Series

A series of the form

[latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots [/latex],

where x is a variable and the coefficients c_n are constants, is known as a power series. The series

[latex]1+x+{x}^{2}+\cdots =\displaystyle\sum _{n=0}^{\infty }{x}^{n}[/latex]

is an example of a power series. Since this series is a geometric series with ratio [latex]r=|x|[/latex], we know that it converges if [latex]|x|<1[/latex] and diverges if [latex]|x|\ge 1[/latex].

Definition

A series of the form

[latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots [/latex]

is a power series centered at [latex]x=0[/latex]. A series of the form

[latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}={c}_{0}+{c}_{1}\left(x-a\right)+{c}_{2}{\left(x-a\right)}^{2}+\cdots [/latex]

is a power series centered at [latex]x=a[/latex].

To make this definition precise, we stipulate that [latex]{x}^{0}=1[/latex] and [latex]{\left(x-a\right)}^{0}=1[/latex] even when [latex]x=0[/latex] and [latex]x=a[/latex], respectively.

The series

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

and

[latex]\displaystyle\sum _{n=0}^{\infty }n\text{!}{x}^{n}=1+x+2\text{!}{x}^{2}+3\text{!}{x}^{3}+\cdots [/latex]

are both power series centered at [latex]x=0[/latex]. The series

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

is a power series centered at [latex]x=2[/latex].

Convergence of a Power Series

Since the terms in a power series involve a variable x, the series may converge for certain values of x and diverge for other values of x. For a power series centered at [latex]x=a[/latex], the value of the series at [latex]x=a[/latex] is given by [latex]{c}_{0}[/latex]. Therefore, a power series always converges at its center. Some power series converge only at that value of x. Most power series, however, converge for more than one value of x. In that case, the power series either converges for all real numbers x or converges for all x in a finite interval. For example, the geometric series [latex]\displaystyle\sum _{n=0}^{\infty }{x}^{n}[/latex] converges for all x in the interval [latex]\left(-1,1\right)[/latex], but diverges for all x outside that interval. We now summarize these three possibilities for a general power series.

theorem: Convergence of a Power Series

Consider the power series [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}[/latex]. The series satisfies exactly one of the following properties:

The series converges at [latex]x=a[/latex] and diverges for all [latex]x\ne a[/latex].
The series converges for all real numbers x.
There exists a real number [latex]R>0[/latex] such that the series converges if [latex]|x-a|<R[/latex] and diverges if [latex]|x-a|>R[/latex]. At the values x where [latex]|x-a|=R[/latex], the series may converge or diverge.

Proof

Suppose that the power series is centered at [latex]a=0[/latex]. (For a series centered at a value of a other than zero, the result follows by letting [latex]y=x-a[/latex] and considering the series [latex]{\displaystyle\sum _{n=1}^{\infty}} {c}_{n} {y}^{n}[/latex].) We must first prove the following fact:

If there exists a real number [latex]d\ne 0[/latex] such that [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{d}^{n}[/latex] converges, then the series [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}[/latex] converges absolutely for all x such that [latex]|x|<|d|[/latex].

Since [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{d}^{n}[/latex] converges, the nth term [latex]{c}_{n}{d}^{n}\to 0[/latex] as [latex]n\to \infty [/latex]. Therefore, there exists an integer N such that [latex]|{c}_{n}{d}^{n}|\le 1[/latex] for all [latex]n\ge N[/latex]. Writing

[latex]|{c}_{n}{x}^{n}|=|{c}_{n}{d}^{n}|{|\frac{x}{d}|}^{n}[/latex],

we conclude that, for all [latex]n\ge N[/latex],

[latex]|{c}_{n}{x}^{n}|\le {|\frac{x}{d}|}^{n}[/latex].

The series

[latex]\displaystyle\sum _{n=N}^{\infty }{|\frac{x}{d}|}^{n}[/latex]

is a geometric series that converges if [latex]|\frac{x}{d}|<1[/latex]. Therefore, by the comparison test, we conclude that [latex]\displaystyle\sum _{n=N}^{\infty }{c}_{n}{x}^{n}[/latex] also converges for [latex]|x|<|d|[/latex]. Since we can add a finite number of terms to a convergent series, we conclude that [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}[/latex] converges for [latex]|x|<|d|[/latex].

With this result, we can now prove the theorem. Consider the series

[latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex]

and let S be the set of real numbers for which the series converges. Suppose that the set [latex]S=\left\{0\right\}[/latex]. Then the series falls under case i. Suppose that the set S is the set of all real numbers. Then the series falls under case ii. Suppose that [latex]S\ne \left\{0\right\}[/latex] and S is not the set of real numbers. Then there exists a real number [latex]x*\ne 0[/latex] such that the series does not converge. Thus, the series cannot converge for any x such that [latex]|x|>|x*|[/latex]. Therefore, the set S must be a bounded set, which means that it must have a smallest upper bound. (This fact follows from the Least Upper Bound Property for the real numbers, which is beyond the scope of this text and is covered in real analysis courses.) Call that smallest upper bound R. Since [latex]S\ne \left\{0\right\}[/latex], the number [latex]R>0[/latex]. Therefore, the series converges for all x such that [latex]|x|<R[/latex], and the series falls into case iii.

[latex]_\blacksquare[/latex]

If a series [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}[/latex] falls into case iii of the Convergence of Power Series, then the series converges for all x such that [latex]|x-a|<R[/latex] for some [latex]R>0[/latex], and diverges for all x such that [latex]|x-a|>R[/latex]. The series may converge or diverge at the values x where [latex]|x-a|=R[/latex]. The set of values x for which the series [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}[/latex] converges is known as the interval of convergence. Since the series diverges for all values x where [latex]|x-a|>R[/latex], the length of the interval is 2R, and therefore, the radius of the interval is R. The value R is called the radius of convergence. For example, since the series [latex]\displaystyle\sum _{n=0}^{\infty }{x}^{n}[/latex] converges for all values x in the interval [latex]\left(-1,1\right)[/latex] and diverges for all values x such that [latex]|x|\ge 1[/latex], the interval of convergence of this series is [latex]\left(-1,1\right)[/latex]. Since the length of the interval is 2, the radius of convergence is 1.

Definition

Consider the power series [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}[/latex]. The set of real numbers x where the series converges is the interval of convergence. If there exists a real number [latex]R>0[/latex] such that the series converges for [latex]|x-a|<R[/latex] and diverges for [latex]|x-a|>R[/latex], then R is the radius of convergence. If the series converges only at [latex]x=a[/latex], we say the radius of convergence is [latex]R=0[/latex]. If the series converges for all real numbers x, we say the radius of convergence is [latex]R=\infty [/latex] (Figure 1).

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

To determine the interval of convergence for a power series, we typically apply the ratio test. In the next example, we show the three different possibilities illustrated in Figure 1.

Recall: Rules for Solving Absolute Value Inequalities

The process of solving an inequality is similar to solving an equation by isolating the variable. There are several rules to keep in mind when solving these inequalities.

The absolute value inequality [latex] |x - a| \le b [/latex] is equivalent to the compound inequality [latex] -b \le x - a \le b [/latex]
Adding or subtracting the same number to both sides of an inequality yields an equivalent statement.
Multiplying or dividing the same positive number to both sides of an inequality yields an equivalent statement.
Multiplying or dividing a negative number to both sides of an inequality reverses the direction of the inequality.
If [latex] |x^n| \le a [/latex], then [latex] -\sqrt[n]{a} \le x \le \sqrt[n] {a} [/latex]

Finally, note that the the solution to an inequality is a range of values that can be expressed using interval notation.

Try It

Example: Finding the Interval and Radius of Convergence

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]

Show Solution

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]-3<x - 2<3[/latex], the series converges absolutely if [latex]-1<x<5[/latex]. The ratio [latex]\rho >1[/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].

try it

Find the interval and radius of convergence for the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{x}^{n}}{\sqrt{n}}[/latex].

Hint

Show Solution

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Module 6: Power Series