Power Series and Functions

A power series is a type of series with terms involving a variable. More specifically, if the variable is x, then all the terms of the series involve powers of x. As a result, a power series can be thought of as an infinite polynomial. Power series are used to represent common functions and also to define new functions. In this section we define power series and show how to determine when a power series converges and when it diverges. We also show how to represent certain functions using power series.

Form of a Power Series

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

r = \| x \|,

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,

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

\sum_{n = 0}^{\infty} x^{n}

but diverges for all x outside that interval. We now summarize these three possibilities for a general power series.

Proof

(For a series centered at a value of a other than zero, the result follows by letting

y = x - a

Therefore, by the comparison test, we conclude that

\sum_{n = N}^{\infty} c_{n} x^{n}

Since we can add a finite number of terms to a convergent series, we conclude that

\sum_{n = 0}^{\infty} c_{n} x^{n}

and let S be the set of real numbers for which the series converges. Suppose that the set

S = {0} .

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

S \neq {0}

and S is not the set of real numbers. Then there exists a real number

x * \neq 0

such that the series does not converge. Thus, the series cannot converge for any x such that

\| x \| > \| x * \| .

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

S \neq {0},

falls into case iii. of [link], then the series converges for all x such that

\| x - a \| < R

The set of values x for which the series

\sum_{n = 0}^{\infty} c_{n} {(x - a)}^{n}

converges is known as the interval of convergence. Since the series diverges for all values x where

\| x - a \| > R,

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

\sum_{n = 0}^{\infty} x^{n}

To determine the interval of convergence for a power series, we typically apply the ratio test. In [link], we show the three different possibilities illustrated in [link].

Finding the Interval and Radius of Convergence

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

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

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

$\begin{matrix} ρ & = lim_{n \to \infty} \| \frac{\frac{x^{n + 1}}{(n + 1)!}}{\frac{x^{n}}{n!}} \| \\ = lim_{n \to \infty} \| \frac{x^{n + 1}}{(n + 1)!} \cdot \frac{n!}{x^{n}} \| \\ = lim_{n \to \infty} \| \frac{x^{n + 1}}{(n + 1) \cdot n!} \cdot \frac{n!}{x^{n}} \| \\ = lim_{n \to \infty} \| \frac{x}{n + 1} \| \\ = \| x \| lim_{n \to \infty} \frac{1}{n + 1} \\ = 0 < 1 \end{matrix}$

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
$R = \infty .$
Apply the ratio test. For $x \neq 0,$
we see that

$\begin{matrix} ρ & = lim_{n \to \infty} \| \frac{(n + 1)! x^{n + 1}}{n! x^{n}} \| \\ = lim_{n \to \infty} \| (n + 1) x \| \\ = \| x \| lim_{n \to \infty} (n + 1) \\ = \infty . \end{matrix}$

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
$x \neq 0 .$
The interval of convergence is the single value
$x = 0$
and the radius of convergence is
$R = 0 .$
In order to apply the ratio test, consider

$\begin{matrix} ρ & = lim_{n \to \infty} \| \frac{\frac{{(x - 2)}^{n + 1}}{(n + 2) 3^{n + 1}}}{\frac{{(x - 2)}^{n}}{(n + 1) 3^{n}}} \| \\ = lim_{n \to \infty} \| \frac{{(x - 2)}^{n + 1}}{(n + 2) 3^{n + 1}} \cdot \frac{(n + 1) 3^{n}}{{(x - 2)}^{n}} \| \\ = lim_{n \to \infty} \| \frac{(x - 2) (n + 1)}{3 (n + 2)} \| \\ = \frac{\| x - 2 \|}{3} . \end{matrix}$

The ratio
$ρ < 1$
if
$\| x - 2 \| < 3 .$
Since
$\| x - 2 \| < 3$
implies that
$−3 < x - 2 < 3,$
the series converges absolutely if
$−1 < x < 5 .$
The ratio
$ρ > 1$
if
$\| x - 2 \| > 3 .$
Therefore, the series diverges if
$x < −1$
or
$x > 5 .$
The ratio test is inconclusive if
$ρ = 1 .$
The ratio
$ρ = 1$
if and only if
$x = −1$
or
$x = 5 .$
We need to test these values of x separately. For
$x = −1,$
the series is given by

$\sum_{n = 0}^{\infty} \frac{{(−1)}^{n}}{n + 1} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ⋯ .$

Since this is the alternating harmonic series, it converges. Thus, the series converges at
$x = −1 .$
For
$x = 5,$
the series is given by

$\sum_{n = 0}^{\infty} \frac{1}{n + 1} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ⋯ .$

This is the harmonic series, which is divergent. Therefore, the power series diverges at
$x = 5 .$
We conclude that the interval of convergence is
$[−1, 5)$
and the radius of convergence is
$R = 3 .$

Representing Functions as Power Series

Being able to represent a function by an “infinite polynomial” is a powerful tool. Polynomial functions are the easiest functions to analyze, since they only involve the basic arithmetic operations of addition, subtraction, multiplication, and division. If we can represent a complicated function by an infinite polynomial, we can use the polynomial representation to differentiate or integrate it. In addition, we can use a truncated version of the polynomial expression to approximate values of the function. So, the question is, when can we represent a function by a power series?

We now show graphically how this series provides a representation for the function

f (x) = \frac{1}{1 - x}

by comparing the graph of f with the graphs of several of the partial sums of this infinite series.

Next we consider functions involving an expression similar to the sum of a geometric series and show how to represent these functions using power series.

In the remaining sections of this chapter, we will show ways of deriving power series representations for many other functions, and how we can make use of these representations to evaluate, differentiate, and integrate various functions.

Key Concepts

Key Equations

In the following exercises, state whether each statement is true, or give an example to show that it is false.

If $\sum_{n = 1}^{\infty} a_{n} x^{n}$

converges, then $a_{n} x^{n} \to 0$

as $n \to \infty .$

True. If a series converges then its terms tend to zero.

\sum_{n = 1}^{\infty} a_{n} x^{n}

converges at $x = 0$

for any real numbers $a_{n} .$

Given any sequence $a_{n},$

there is always some $R > 0,$

possibly very small, such that $\sum_{n = 1}^{\infty} a_{n} x^{n}$

converges on $(− R, R) .$

False. It would imply that $a_{n} x^{n} \to 0$

for $\| x \| < R .$

If $a_{n} = n^{n},$

then $a_{n} x^{n} = {(n x)}^{n}$

does not tend to zero for any $x \neq 0 .$

If $\sum_{n = 1}^{\infty} a_{n} x^{n}$

has radius of convergence $R > 0$

and if $\| b_{n} \| \leq \| a_{n} \|$

for all n, then the radius of convergence of $\sum_{n = 1}^{\infty} b_{n} x^{n}$

is greater than or equal to R.

Suppose that $\sum_{n = 0}^{\infty} a_{n} {(x - 3)}^{n}$

converges at $x = 6 .$

At which of the following points must the series also converge? Use the fact that if $\sum a_{n} {(x - c)}^{n}$

converges at x, then it converges at any point closer to c than x.

$x = 1$
$x = 2$
$x = 3$
$x = 0$
$x = 5.99$
$x = 0.000001$

It must converge on $(0, 6]$

and hence at: a. $x = 1;$

b. $x = 2;$

c. $x = 3;$

d. $x = 0;$

e. $x = 5.99;$

and f. $x = 0.000001 .$

Suppose that $\sum_{n = 0}^{\infty} a_{n} {(x + 1)}^{n}$

converges at $x = −2 .$

At which of the following points must the series also converge? Use the fact that if $\sum a_{n} {(x - c)}^{n}$

converges at x, then it converges at any point closer to c than x.

$x = 2$
$x = −1$
$x = −3$
$x = 0$
$x = 0.99$
$x = 0.000001$

In the following exercises, suppose that $\| \frac{a_{n + 1}}{a_{n}} \| \to 1$

as $n \to \infty .$

Find the radius of convergence for each series.

\sum_{n = 0}^{\infty} a_{n} 2^{n} x^{n}

\| \frac{a_{n + 1} 2^{n + 1} x^{n + 1}}{a_{n} 2^{n} x^{n}} \| = 2 \| x \| \| \frac{a_{n + 1}}{a_{n}} \| \to 2 \| x \|

so $R = \frac{1}{2}$

\sum_{n = 0}^{\infty} \frac{a_{n} x^{n}}{2^{n}}

\sum_{n = 0}^{\infty} \frac{a_{n} π^{n} x^{n}}{e^{n}}

\| \frac{a_{n + 1} {(\frac{π}{e})}^{n + 1} x^{n + 1}}{a_{n} {(\frac{π}{e})}^{n} x^{n}} \| = \frac{π \| x \|}{e} \| \frac{a_{n + 1}}{a_{n}} \| \to \frac{π \| x \|}{e}

so $R = \frac{e}{π}$

\sum_{n = 0}^{\infty} \frac{a_{n} {(−1)}^{n} x^{n}}{10^{n}}

\sum_{n = 0}^{\infty} a_{n} {(−1)}^{n} x^{2 n}

\| \frac{a_{n + 1} {(−1)}^{n + 1} x^{2 n + 2}}{a_{n} {(−1)}^{n} x^{2 n}} \| = \| x^{2} \| \| \frac{a_{n + 1}}{a_{n}} \| \to \| x^{2} \|

so $R = 1$

\sum_{n = 0}^{\infty} a_{n} {(−4)}^{n} x^{2 n}

In the following exercises, find the radius of convergence R and interval of convergence for $\sum a_{n} x^{n}$

with the given coefficients $a_{n} .$

\sum_{n = 1}^{\infty} \frac{{(2 x)}^{n}}{n}

a_{n} = \frac{2^{n}}{n}

so $\frac{a_{n + 1} x}{a_{n}} \to 2 x .$

so $R = \frac{1}{2} .$

When $x = \frac{1}{2}$

the series is harmonic and diverges. When $x = - \frac{1}{2}$

the series is alternating harmonic and converges. The interval of convergence is $I = [- \frac{1}{2}, \frac{1}{2}) .$

\sum_{n = 1}^{\infty} {(−1)}^{n} \frac{x^{n}}{\sqrt{n}}

\sum_{n = 1}^{\infty} \frac{n x^{n}}{2^{n}}

a_{n} = \frac{n}{2^{n}}

so $\frac{a_{n + 1} x}{a_{n}} \to \frac{x}{2}$

so $R = 2 .$

When $x = ± 2$

the series diverges by the divergence test. The interval of convergence is $I = (−2, 2) .$

\sum_{n = 1}^{\infty} \frac{n x^{n}}{e^{n}}

\sum_{n = 1}^{\infty} \frac{n^{2} x^{n}}{2^{n}}

a_{n} = \frac{n^{2}}{2^{n}}

so $R = 2 .$

When $x = ± 2$

the series diverges by the divergence test. The interval of convergence is $I = (−2, 2) .$

\sum_{k = 1}^{\infty} \frac{k^{e} x^{k}}{e^{k}}

\sum_{k = 1}^{\infty} \frac{π^{k} x^{k}}{k^{π}}

a_{k} = \frac{π^{k}}{k^{π}}

so $R = \frac{1}{π} .$

When $x = ± \frac{1}{π}$

the series is an absolutely convergent p-series. The interval of convergence is $I = [- \frac{1}{π}, \frac{1}{π}] .$

\sum_{n = 1}^{\infty} \frac{x^{n}}{n!}

\sum_{n = 1}^{\infty} \frac{10^{n} x^{n}}{n!}

a_{n} = \frac{10^{n}}{n!}, \frac{a_{n + 1} x}{a_{n}} = \frac{10 x}{n + 1} \to 0 < 1

so the series converges for all x by the ratio test and $I = (− \infty, \infty) .$

\sum_{n = 1}^{\infty} {(−1)}^{n} \frac{x^{n}}{ln (2 n)}

In the following exercises, find the radius of convergence of each series.

\sum_{k = 1}^{\infty} \frac{{(k!)}^{2} x^{k}}{(2 k)!}

a_{k} = \frac{{(k!)}^{2}}{(2 k)!}

so $\frac{a_{k + 1}}{a_{k}} = \frac{{(k + 1)}^{2}}{(2 k + 2) (2 k + 1)} \to \frac{1}{4}$

so $R = 4$

\sum_{n = 1}^{\infty} \frac{(2 n)! x^{n}}{n^{2 n}}

\sum_{k = 1}^{\infty} \frac{k!}{1 \cdot 3 \cdot 5 ⋯ (2 k - 1)} x^{k}

a_{k} = \frac{k!}{1 \cdot 3 \cdot 5 ⋯ (2 k - 1)}

so $\frac{a_{k + 1}}{a_{k}} = \frac{k + 1}{2 k + 1} \to \frac{1}{2}$

so $R = 2$

\sum_{k = 1}^{\infty} \frac{2 \cdot 4 \cdot 6 ⋯ 2 k}{(2 k)!} x^{k}

\sum_{n = 1}^{\infty} \frac{x^{n}}{(\begin{matrix} 2 n \\ n \end{matrix})}

where $(\begin{matrix} n \\ k \end{matrix}) = \frac{n!}{k! (n - k)!}$

a_{n} = \frac{1}{(\begin{matrix} 2 n \\ n \end{matrix})}

so $\frac{a_{n + 1}}{a_{n}} = \frac{{((n + 1)!)}^{2}}{(2 n + 2)!} \frac{2 n!}{{(n!)}^{2}} = \frac{{(n + 1)}^{2}}{(2 n + 2) (2 n + 1)} \to \frac{1}{4}$

so $R = 4$

\sum_{n = 1}^{\infty} {sin}^{2} n x^{n}

In the following exercises, use the ratio test to determine the radius of convergence of each series.

\sum_{n = 1}^{\infty} \frac{{(n!)}^{3}}{(3 n)!} x^{n}

\frac{a_{n + 1}}{a_{n}} = \frac{{(n + 1)}^{3}}{(3 n + 3) (3 n + 2) (3 n + 1)} \to \frac{1}{27}

so $R = 27$

\sum_{n = 1}^{\infty} \frac{2^{3 n} {(n!)}^{3}}{(3 n)!} x^{n}

\sum_{n = 1}^{\infty} \frac{n!}{n^{n}} x^{n}

a_{n} = \frac{n!}{n^{n}}

so $\frac{a_{n + 1}}{a_{n}} = \frac{(n + 1)!}{n!} \frac{n^{n}}{{(n + 1)}^{n + 1}} = {(\frac{n}{n + 1})}^{n} \to \frac{1}{e}$

so $R = e$

\sum_{n = 1}^{\infty} \frac{(2 n)!}{n^{2 n}} x^{n}

In the following exercises, given that $\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^{n}$

with convergence in $(−1, 1),$

find the power series for each function with the given center a, and identify its interval of convergence.

f (x) = \frac{1}{x}; a = 1

(Hint: $\frac{1}{x} = \frac{1}{1 - (1 - x)})$

f (x) = \sum_{n = 0}^{\infty} {(1 - x)}^{n}

on $I = (0, 2)$

f (x) = \frac{1}{1 - x^{2}}; a = 0

f (x) = \frac{x}{1 - x^{2}}; a = 0

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

on $I = (−1, 1)$

f (x) = \frac{1}{1 + x^{2}}; a = 0

f (x) = \frac{x^{2}}{1 + x^{2}}; a = 0

\sum_{n = 0}^{\infty} {(−1)}^{n} x^{2 n + 2}

on $I = (−1, 1)$

f (x) = \frac{1}{2 - x}; a = 1

f (x) = \frac{1}{1 - 2 x}; a = 0 .

\sum_{n = 0}^{\infty} 2^{n} x^{n}

on $(- \frac{1}{2}, \frac{1}{2})$

f (x) = \frac{1}{1 - 4 x^{2}}; a = 0

f (x) = \frac{x^{2}}{1 - 4 x^{2}}; a = 0

\sum_{n = 0}^{\infty} 4^{n} x^{2 n + 2}

on $(- \frac{1}{2}, \frac{1}{2})$

f (x) = \frac{x^{2}}{5 - 4 x + x^{2}}; a = 2

Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.

Explain why, if ${\| a_{n} \|}^{1 / n} \to r > 0,$

then ${\| a_{n} x^{n} \|}^{1 / n} \to \| x \| r < 1$

whenever $\| x \| < \frac{1}{r}$

and, therefore, the radius of convergence of $\sum_{n = 1}^{\infty} a_{n} x^{n}$

is $R = \frac{1}{r} .$

{\| a_{n} x^{n} \|}^{1 / n} = {\| a_{n} \|}^{1 / n} \| x \| \to \| x \| r

as $n \to \infty$

and $\| x \| r < 1$

when $\| x \| < \frac{1}{r} .$

Therefore, $\sum_{n = 1}^{\infty} a_{n} x^{n}$

converges when $\| x \| < \frac{1}{r}$

by the nth root test.

\sum_{n = 1}^{\infty} \frac{x^{n}}{n^{n}}

\sum_{k = 1}^{\infty} {(\frac{k - 1}{2 k + 3})}^{k} x^{k}

a_{k} = {(\frac{k - 1}{2 k + 3})}^{k}

so ${(a_{k})}^{1 / k} \to \frac{1}{2} < 1$

so $R = 2$

\sum_{k = 1}^{\infty} {(\frac{2 k^{2} - 1}{k^{2} + 3})}^{k} x^{k}

\sum_{n = 1}^{\infty} a_{n} = {(n^{1 / n} - 1)}^{n} x^{n}

a_{n} = {(n^{1 / n} - 1)}^{n}

so ${(a_{n})}^{1 / n} \to 0$

so $R = \infty$

Suppose that $p (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$

such that $a_{n} = 0$

if n is even. Explain why $p (x) = - p (− x) .$

Suppose that $p (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$

such that $a_{n} = 0$

if n is odd. Explain why $p (x) = p (− x) .$

We can rewrite $p (x) = \sum_{n = 0}^{\infty} a_{2 n + 1} x^{2 n + 1}$

and $p (x) = p (− x)$

since $x^{2 n + 1} = − {(− x)}^{2 n + 1} .$

Suppose that $p (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$

converges on $(−1, 1] .$

Find the interval of convergence of $p (A x) .$

Suppose that $p (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$

converges on $(−1, 1] .$

Find the interval of convergence of $p (2 x - 1) .$

If $x \in [0, 1],$

then $y = 2 x - 1 \in [−1, 1]$

so $p (2 x - 1) = p (y) = \sum_{n = 0}^{\infty} a_{n} y^{n}$

converges.

In the following exercises, suppose that $p (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$

satisfies $lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = 1$

where $a_{n} \geq 0$

for each n. State whether each series converges on the full interval $(−1, 1),$

or if there is not enough information to draw a conclusion. Use the comparison test when appropriate.

\sum_{n = 0}^{\infty} a_{n} x^{2 n}

\sum_{n = 0}^{\infty} a_{2 n} x^{2 n}

Converges on $(−1, 1)$

by the ratio test

\sum_{n = 0}^{\infty} a_{2 n} x^{n} (H i n t : x = ± \sqrt{x^{2}})

\sum_{n = 0}^{\infty} a_{n^{2}} x^{n^{2}}

(Hint: Let $b_{k} = a_{k}$

if $k = n^{2}$

for some n, otherwise $b_{k} = 0 .)$

Consider the series $\sum b_{k} x^{k}$

where $b_{k} = a_{k}$

if $k = n^{2}$

and $b_{k} = 0$

otherwise. Then $b_{k} \leq a_{k}$

and so the series converges on $(−1, 1)$

by the comparison test.

Suppose that $p (x)$

is a polynomial of degree N. Find the radius and interval of convergence of $\sum_{n = 1}^{\infty} p (n) x^{n} .$

[T] Plot the graphs of $\frac{1}{1 - x}$

and of the partial sums $S_{N} = \sum_{n = 0}^{N} x^{n}$

for $n = 10, 20, 30$

on the interval $[−0.99, 0.99] .$

Comment on the approximation of $\frac{1}{1 - x}$

by $S_{N}$

near $x = −1$

and near $x = 1$

as N increases.

This figure is the graph of y = 1/(1-x), which is an increasing curve with vertical asymptote at 1.

The approximation is more accurate near $x = −1 .$

The partial sums follow $\frac{1}{1 - x}$

more closely as N increases but are never accurate near $x = 1$

since the series diverges there.

[T] Plot the graphs of $− ln (1 - x)$

and of the partial sums $S_{N} = \sum_{n = 1}^{N} \frac{x^{n}}{n}$

for $n = 10, 50, 100$

on the interval $[−0.99, 0.99] .$

Comment on the behavior of the sums near $x = −1$

and near $x = 1$

as N increases.

[T] Plot the graphs of the partial sums $S_{n} = \sum_{n = 1}^{N} \frac{x^{n}}{n^{2}}$

for $n = 10, 50, 100$

on the interval $[−0.99, 0.99] .$

Comment on the behavior of the sums near $x = −1$

and near $x = 1$

as N increases.

This figure is the graph of y = -ln(1-x) which is an increasing curve passing through the origin.

The approximation appears to stabilize quickly near both $x = ± 1 .$

[T] Plot the graphs of the partial sums $S_{N} = \sum_{n = 1}^{N} sin n x^{n}$

for $n = 10, 50, 100$

on the interval $[−0.99, 0.99] .$

Comment on the behavior of the sums near $x = −1$

and near $x = 1$

as N increases.

[T] Plot the graphs of the partial sums $S_{N} = \sum_{n = 0}^{N} {(−1)}^{n} \frac{x^{2 n + 1}}{(2 n + 1)!}$

for $n = 3, 5, 10$

on the interval $[−2 π, 2 π] .$

Comment on how these plots approximate $sin x$

as N increases.

This figure is the graph of the partial sums of (-1)^n times x^(2n+1) divided by (2n+1)! For n=3,5,10. The curves approximate the sine curve close to the origin and then separate as the curves move away from the origin.

The polynomial curves have roots close to those of $sin x$

up to their degree and then the polynomials diverge from $sin x .$

[T] Plot the graphs of the partial sums $S_{N} = \sum_{n = 0}^{N} {(−1)}^{n} \frac{x^{2 n}}{(2 n)!}$

for $n = 3, 5, 10$

on the interval $[−2 π, 2 π] .$

Comment on how these plots approximate $cos x$

as N increases.

Power Series and Functions

Form of a Power Series

Convergence of a Power Series

Proof

Representing Functions as Power Series

Key Concepts

Key Equations

Glossary