### 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

where *x* is a variable and the coefficients *c _{n}* are constants, is known as a

**power series**. The series

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

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

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

and

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

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 *n*th 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

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

The series

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

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 2*R*, 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).

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]

### try it

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

Watch the following video to see the worked solution to the above Try It.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “6.1.2” here (opens in new window).