Lesson 17 - Power Series

Form of a Power Series

A series of the form

\sum_{n = 0}^{\infty} c_{n} x^{n} = C_{0} + C_{1} x + c_{2} x^{2} + \dots,

where $x$ is a variable and the coefficients $C_{n}$ are constants is known as a power series. The series

1 + x + x^{2} + \dots = \sum_{n = 0}^{\infty} x^{n}

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

A series of the form

\sum_{n = 1}^{\infty} C_{n} x^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots

is a power series centered at $x = 0$ . A series of the form

\sum_{n = 0}^{\infty} C_{n} (x - a)^{n} = C_{0} + C_{1} (x - a) + C_{2} (x - a)^{2} + \dots

is a power series centered at $x = a$ .

It is important to note that $x^{0} = 1$ and $(x - a)^{0} = 1$ even when $x = 0$ and $x = a$ , respectively.

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 $x = a$ , then value of the series at $x = a$ is given by $C_{0}$ . Therefore, a power series always converges at its center. Most power series converge for more than one value of $x$ , either for all real numbers $x$ or for all $x$ in a finite interval. For this, we turn to a theorem.

Theorem: Convergence of a Power Series

Consider the power series $\sum_{n = 0}^{\infty} C_{n} (x - a)^{n}$ . The series satisfies exactly one of the following properties:
i. The series converges at $x = a$ and diverges for all $x \neq a$ .
ii. The series converges for all real numbers $x$ .
iii. There exists a real number $R > 0$ . At the values $x$ where $| x - a | = R$ , the series may converge or diverge.

A proof for this theorem can be found here.

Interval of Convergence and Radius of Convergence

Consider the power series $\sum_{n = 0}^{\infty} C_{n} (x - a)^{n}$ . The set of real numbers $x$ where the series converges is the interval of convergence. If there exists a real number $R > 0$ then $R$ is the radius of convergence. If the series converges only at $x = a$ , we say the radius of convergence is $R = 0$ . If the series converges for all real numbers $x$ , we say the radius of convergence is $R = \infty$ .

Example: Finding the Interval and Radius of Convergence

a. $\sum_{n = 0}^{\infty} \frac{x^{n}}{n!}$
b. $\sum_{n = 0}^{\infty} n! x^{n}$
c. $\sum_{n = 0}^{\infty} \frac{(x - 2)^{n}}{(n + 1) 3^{n}}$

a. The check for convergence, apply the ratio test. We have

\begin{array}{l} ρ = lim_{n \to \infty} | \frac{\frac{x^{n + 1}}{(n + 1)!}}{\frac{x^{n}}{n!}} | \\ = lim_{n \to \infty} | \frac{x^{n + 1}}{(n + 1)!} \times \frac{n!}{x^{n}} | \\ = lim_{n \to \infty} | \frac{x^{n + 1}}{(n + 1) \times n!} \times \frac{n!}{x^{n}} | \\ = lim_{n \to \infty} | \frac{x}{n + 1} | \\ = | x | lim_{n \to \infty} \frac{1}{n + 1} \\ = 0 < 1 \end{array}

for all values of $x$ . Therefore, the series converges for all real numbers $x$ . The interval of convergence is $(- \infty, \infty)$ and the radius of convergence is $R = \infty$ .

b. Apply the ratio test. For $x \neq 0$ , we see that

\begin{array}{l} ρ = lim_{n \to \infty} | \frac{(n + 1)! x^{n + 1}}{n! x^{n}} | \\ = lim_{n \to \infty} | (n + 1) x | \\ = | x | lim_{n \to \infty} (n + 1) \\ = \infty \end{array}

Therefore, the series diverges for all $x \neq 0$ . Since the series is centered at $x = 0$ , it must converge there, so the series converges only for $x = 0$ . The interval of convergence is the single value $x = 0$ and the radius of convergence is $R = 0$ .

c. Try for yourself.

Representing Functions as Power Series

Consider again the geometric series

1 + x + x^{2} + x^{3} + \dots = \sum_{n = 0}^{\infty} x^{n}

Recall that the geometric series

a + a r + a r^{2} + a r^{3} + \dots

converges if and only if $| r | < 1$ .

In that case, it converges to $\frac{a}{1 - r} .$ Therefore, if $| x | < 1$ , the series in the example above converges to $\frac{1}{1 - x}$ and we write

1 + x + x^{2} + x^{3} + \dots = \frac{1}{1 - x} f o r | x | < 1

As a result, we are able to represent the function $f (x) = \frac{1}{1 - x}$ for $| x | < 1$ by the power series

1 + x + x^{2} + x^{3} + \dots w h e n | x | < 1.

Example: Representing a Function with a Power Series

Use a power series to represent each of the following functions $f$ . Find the interval of convergence.

a. $f (x) = \frac{1}{1 + x^{3}}$
b. $f (x) = \frac{x^{2}}{4 - x^{2}}$

a. You should recognize this function $f$ as the sum of a geometric series, because

\frac{1}{1 + x^{3}} = \frac{1}{1 - (- x^{3})} .

Using the fact that, for $| r | < 1$ , $\frac{a}{1 - r}$ is the sum of the geometric series

\sum_{n = 0}^{\infty} a r^{n} = a + a r + a r^{2} + \dots,

we see that, for $| - x^{3} | < 1$ ,

\begin{array}{l} \frac{1}{1 + x^{3}} = \frac{1}{1 - (- x^{3})} \\ = \sum_{n = 0}^{\infty} (- x^{3})^{n} \\ = 1 - x^{3} + x^{6} - x^{9} + \dots \end{array}

Since this series converges if and only if $| - x^{3} | < 1$ , the interval of convergence is $(- 1, 1)$ , and we have

\frac{1}{1 + x^{3}} = 1 - x^{3} + x^{6} - x^{9} + \dots f o r | x | < 1.

b. Try for yourself.

Next Lesson: Lesson 18 - Taylor Series

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Note: This lesson adapted from OpenStax. Credit goes to OpenStax creators.