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## Section7.5Power Series

###### Motivating Questions
• What is a power series?

• What are some important uses of power series?

• What is the interval of convergence? For what values does a power series converge?

So far, each infinite series we have discussed has been a series of real numbers, such as

\begin{equation} 1 + \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^k} + \cdots = \sum_{k=0}^{\infty} \frac{1}{2^k}\text{.}\label{E-geom12}\tag{7.18} \end{equation}

In the remainder of this chapter, we will include series that involve a variable. For instance, if in the geometric series in Equation(7.18) we replace the ratio $r = \frac{1}{2}$ with the variable $x\text{,}$ we have the infinite (still geometric) series

\begin{equation} 1 + x + x^2 + \cdots + x^k + \cdots = \sum_{k=0}^{\infty} x^k\text{.}\label{E-geomx}\tag{7.19} \end{equation}

In this section we will focus on more general series of powers. As a motivation, consider the following example.

###### Example7.46

Consider the power series defined by

\begin{equation*} f(x) = \sum_{k=0}^{\infty} \frac{x^k}{2^k}\text{.} \end{equation*}

What are $f(1)$ and $f\left(\frac{3}{2}\right)\text{?}$ Find a general formula for $f(x)$ and determine the values for which this power series converges.

Solution

If we evaluate $f$ at $x=1$ we obtain the series

\begin{equation*} \sum_{k=0}^{\infty} \frac{1}{2^k} \end{equation*}

which is a geometric series with ratio $\frac{1}{2}\text{.}$ So we can sum this series and find that

\begin{equation*} f(1) = \frac{1}{1-\frac{1}{2}} = 2\text{.} \end{equation*}

Similarly,

\begin{equation*} f(3/2) = \sum_{k=0}^{\infty} \left(\frac{3}{4}\right)^k = \frac{1}{1-\frac{3}{4}} = 4\text{.} \end{equation*}

In general, $f(x)$ is a geometric series with ratio $\frac{x}{2}\text{,}$ so

\begin{equation*} f(x) = \sum_{k=0}^{\infty} \left(\frac{x}{2}\right)^k = \frac{1}{1-\frac{x}{2}} = \frac{2}{2-x} \end{equation*}

provided that $-1 \lt \frac{x}{2} \lt 1$ (which ensures that the ratio is less than 1 in absolute value). Thus, the power series that defines $f(x)=\frac{2}{2-x}$ converges for $-2 \lt x \lt 2\text{.}$

As Example7.46 illustrates, a power series may converge for some values of $x$ and not for others. In this section, we will learn how to determine the interval of $x$-values where a power series converges. In the following sections we will show how power series may be used to obtain polynomial approximations of functions.

### SubsectionPower Series

###### Power Series

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

\begin{equation} \sum_{k=0}^{\infty} c_k(x-a)^k\label{eq-8-6-power-series}\tag{7.20} \end{equation}

where $\{c_k\}$ is a sequence of real numbers and $x$ is an independent variable.

For example, the series

\begin{equation*} 1 + (x-2) + (x-2)^2 + (x-2)^3 + \cdots \end{equation*}

is a power series centered at $x = 2$ with $c_i = 1$ for $i \ge 1 \text{,}$ and a geometric series

\begin{equation*} \sum_{n=1}^{\infty} bx^n = b + bx + bx^2 + bx^3 + \cdots = \sum_{n=0}^{\infty} b(x-0)^n \end{equation*}

is a power series centered at $x=0$ with $c_i = b$ for $i \ge 1 \text{.}$

Convergence of power series is similar to convergence of series. Namely, a power series will converge if its sequence of partial sums converges. In general, a power series may converge for some values of $x \text{,}$ and diverge for others.

###### Convergence of Power Series

For a fixed value of $x \text{,}$ the power series

\begin{equation*} \sum_{k=0}^{\infty} c_k(x-a)^k \end{equation*}

converges to $L$ if the sequence of partial sums $S_0(x), S_1(x), S_2(x), \ldots$ converges to $L \text{.}$ That is,

\begin{equation*} \lim_{n \rightarrow \infty} S_n(x) = L \text{.} \end{equation*}

The set of $x$ values at which a power series $\sum_{k=0}^{\infty} c_k(x-a)^k$ converges is always an interval centered at $x=a\text{.}$ For this reason, the set on which a power series converges is called the interval of convergence. Half the length of the interval of convergence is called the radius of convergence.

###### Example7.47

Consider the series $\sum_{n=0}^{\infty} \frac{ (x-1)^n}{2^n} \text{.}$ This is a power series about $x = 1 \text{.}$

1. Does the series converge or diverge when $x =2 \text{?}$

2. Does it converge or diverge when $x = 3 \text{?}$

3. For what values of $x$ will the series converge?

1. The series converges at $x=2\text{.}$

2. It diverges at $x=3\text{.}$

3. The series converges for values of $x$ satisfying $-1 \lt x \lt 3 \text{.}$

Solution
1. To test, we can substitute in $x=2$ into the series, giving:

\begin{equation*} \sum_{n=0}^\infty \frac{(2-1)^n}{2^n} = \sum_{n=0}^\infty \frac{1}{2^n}. \end{equation*}

This series is geometric with ratio $\frac{1}{2}\text{,}$ so it converges (in fact, to 2).

2. We make the substitution $x=3$ into the series, and get

\begin{equation*} \sum_{n=0}^\infty \frac{(3-1)^n}{2^n} = \sum_{n=0}^\infty \frac{2^n}{2^n} = \sum_{n=0}^\infty 1 = 1 + 1 + 1 + \cdots\text{.} \end{equation*}

Since this series is unbounded, it diverges.

3. To figure out exactly when the series converges, we'll start by using the Ratio Test. Here $a_n= \frac{(x-1)^n}{2^n}\text{.}$ So, notice

\begin{equation*} \left|\frac{a_{n+1}}{a_n}\right| = \frac{|x-1|^{n+1}}{2^{n+1}} \cdot \frac{2^n}{|x-1|^n} = \frac{|x-1|}{2}. \end{equation*}

Thus, the series converges whenever $\frac{|x-1|}{2}\lt 1$ and diverges whenever $\frac{|x-1|}{2} \gt 1 \text{.}$ Examining that first inequality yields:

\begin{align*} \frac{|x-1|}{2} \amp \lt 1 \\ |x-1| \amp \lt 2 \\ -2 \amp\lt x-1 \lt 2 \\ -1 \amp\lt x \lt 3 \end{align*}

So now we know that the series converges when $-1 \lt x \lt 3$ and diverges when $x \lt -1$ and when $x \gt 3 \text{.}$ Unfortunately, the ratio test can't tell us what happens exactly at $x=-1$ or at $x=3\text{,}$ because that's when $\left|\frac{a_{n+1}}{a_n}\right|=1\text{.}$ However, we can just check those directly. We determined in part b. that the series diverges at $x=3\text{,}$ so we just need to check $x=-1\text{:}$

\begin{equation*} \sum_{n=0}^\infty \frac{(-1-1)^n}{2^n}= \sum_{n=0}^\infty \frac{(-2)^n}{2^n} = \sum_{n=0}^\infty -1 = -1 -1 -1 \cdots\text{.} \end{equation*}

This series is unbounded below, so it diverges.

Thus, the series converges exactly when $-1 \lt x \lt 3$ and diverges everywhere else.

### SubsectionFinding the Interval of Convergence

A power series defines a function $f$ whose domain is the set of $x$ values for which the power series converges. We therefore write

\begin{equation*} f(x) = \sum_{k=0}^{\infty} c_k(x-a)^k\text{.} \end{equation*}

To determine the values of $x$ for which a power series

\begin{equation*} \sum_{k=0}^{\infty} c_k (x-a)^k\text{,} \end{equation*}

centered at $x = a$ will converge, we apply the Ratio Test with $a_k = | c_k (x-a)^k |\text{.}$ The series converges if $\lim_{k \to \infty} \frac{a_{k+1}}{a_k} \lt 1\text{.}$

Observe that

\begin{equation*} \frac{a_{k+1}}{{a_k}} = | x-a | \frac{| c_{k+1} |}{| c_{k} |}\text{,} \end{equation*}

so when we apply the Ratio Test, we get

\begin{equation*} \lim_{k \to \infty} \frac{a_{k+1}}{a_k} = \lim_{k \to \infty} |x-a| \frac{| c_{k+1} |}{| c_{k} |}\text{.} \end{equation*}

Note suppose that

\begin{equation*} \lim_{k \to \infty} \frac{| c_{k+1} |}{| c_{k} |} = L\text{,} \end{equation*}

so that

\begin{equation*} \lim_{k \to \infty} \frac{a_{k+1}}{a_k} = |x-a| \cdot L\text{.} \end{equation*}

There are three possibilities for $L\text{:}$ $L$ can be $0\text{,}$ it can be a finite positive value, or it can be infinite. Based on this value of $L\text{,}$ we can determine for which values of $x$ the original power series converges.

• If $L = 0\text{,}$ then the power series converges on $(-\infty, \infty)\text{.}$

• If $L$ is infinite, then the power series converges only at $x = a\text{.}$

• If $L$ is finite and nonzero, then the power series converges absolutely for all $x$ that satisfy

\begin{equation*} |x-a| \cdot L \lt 1 \end{equation*}

or for all $x$ such that

\begin{equation*} |x-a| \lt \frac{1}{L}\text{,} \end{equation*}

which is the interval

\begin{equation*} \left(a-\frac{1}{L}, a+\frac{1}{L}\right)\text{.} \end{equation*}

This interval is centered at $a$ and has radius $R = \frac{1}{L} \text{.}$ Because the Ratio Test is inconclusive when the $|x-a| \cdot L = 1\text{,}$ the endpoints $a \pm \frac{1}{L}$ have to be checked separately.

###### Finding the Interval of Convergence

To find the interval of convergence of the power series $\sum_{k=0}^{\infty} c_k(x-a)^k \text{,}$ apply the ratio test to obtain

\begin{equation*} \lim_{k \rightarrow \infty} |x - a|\frac{|c_{k+1}|}{|c_k|} = |x - a| \cdot L \end{equation*}

There are three cases.

• If $L = 0 \text{,}$ then the power series converges for all values of $x \text{.}$ Thus, the the interval of convergence is $(-\infty, +\infty)$ and the radius of convergence is $R = \infty \text{.}$

• If $L = \infty \text{,}$ then the power series converges only at $x = a \text{,}$ and the radius of convergence is $R = 0 \text{.}$

• If $L$ is finite and nonzero, then the power series converges for all $x \in (a - \frac{1}{L}, a + \frac{1}{L}) \text{,}$ so the radius of convergence is $R = \frac{1}{L} \text{.}$ The power series may or may not converge at each endpoint, so testing for convergence at $x = a - R$ and $x = a + R$ is necessary.

###### Example7.48

Let $f(x) = \sum_{k=1}^{\infty} \frac{x^k}{k^2}\text{.}$ Determine the interval of convergence of this power series.

Solution

First we will plot some of the partial sums of this power series to get an idea of the interval of convergence. Let

\begin{equation*} S_n(x) = \sum_{k=1}^{n} \frac{x^k}{k^2} \end{equation*}

for each $n \geq 1\text{.}$ Figure7.49 shows plots of $S_{10}(x)$ (in red), $S_{25}(x)$ (in blue), and $S_{50}(x)$ (in green). Figure7.49Graphs of some partial sums of the power series $\sum_{k=1}^{\infty} \frac{x^k}{k^2}\text{.}$

The behavior of $S_{50}$ in particular suggests that $f(x)$ appears to be converging to a particular curve on the interval $(-1,1)\text{,}$ while growing without bound outside of that interval. Thus, the interval of convergence might be $-1 \lt x \lt 1\text{.}$ To verify our conjecture, we apply the Ratio Test. Now,

\begin{equation*} a_k = \frac{x^k}{k^2}\text{,} \end{equation*}

so

\begin{align*} \lim_{k \to \infty} \frac{\left| a_{k+1} \right|}{ \left| a_k \right|} \amp = \lim_{k \to \infty} \frac{ \frac{|x|^{k+1}}{(k+1)^2} }{ \frac{| x|^{k}}{k^2} }\\ \amp = \lim_{k \to \infty} |x| \left(\frac{k}{k+1}\right)^2\\ \amp = |x| \lim_{k \to \infty} \left(\frac{k}{k+1}\right)^2\\ \amp = |x|\text{.} \end{align*}

Therefore, the Ratio Test tells us that $f(x)$ converges absolutely when $| x | \lt 1$ and diverges when $| x | \gt 1\text{.}$ Because the Ratio Test is inconclusive when $|x| = 1\text{,}$ we need to check $x = 1$ and $x = -1$ individually.

When $x = 1\text{,}$ observe that

\begin{equation*} f(1) = \sum_{k=1}^{\infty} \frac{1}{k^2}\text{.} \end{equation*}

This is a $p$-series with $p \gt 1\text{,}$ which we know converges. When $x = -1\text{,}$ we have

\begin{equation*} f(-1) = \sum_{k=1}^{\infty} \frac{(-1)^k}{k^2}\text{.} \end{equation*}

This is an alternating series, and since the sequence $\left\{ \frac{1}{n^2} \right\}$ decreases to 0, the power series converges by the Alternating Series Test. Thus, the interval of convergence of this power series is $-1 \le x \le 1\text{.}$

###### Example7.50

Determine the interval of convergence of each power series.

1. $\sum_{k=1}^{\infty} \frac{(x-1)^k}{3k}$

2. $\sum_{k=1}^{\infty} kx^k$

3. $\sum_{k=1}^{\infty} \frac{k^2(x+1)^k}{4^k}$

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

5. $\sum_{k=1}^{\infty} k!x^k$

1. $[0,2)\text{.}$

2. $(-1,1)\text{.}$

3. $(-5,3)\text{.}$

4. $(-\infty, \infty)\text{.}$

5. $\{0\}\text{.}$

Solution
1. We use the Ratio Test with $a_k = \frac{|x-1|^k}{3k}\text{:}$

\begin{align*} \lim_{k \to \infty} \frac{ \frac{|x-1|^{k+1}}{3(k+1)} }{ \frac{|x-1|^k}{3k} } \amp = \lim_{k \to \infty} \frac{3k|x-1|^{k+1}}{3( k+1)|x-1|^{k}}\\ \amp = |x-1| \lim_{k \to \infty} \frac{k}{k+1}\\ \amp = |x-1|\text{.} \end{align*}

So the power series $\sum_{k=1}^{\infty} \frac{(x-1)^k}{3k}$ converges absolutely when $|x-1| \lt 1$ or when $0 \lt x \lt 2$ and diverges outside this interval. To completely determine the interval of convergence, we need to check what happens at the endpoints of this interval.

• When $x=0$ our power series is $\sum_{k=1}^{\infty} \frac{(-1)^k}{3k}$ which is just a scalar multiple of the alternating harmonic series and so converges.

• When $x=2$ our power series is $\sum_{k=1}^{\infty} \frac{1}{3k}$ which is just a scalar multiple of the harmonic series and so diverges.

Therefore, the interval of convergence of the power series $\sum_{k=1}^{\infty} \frac{(x-1)^k}{3k}$ is $[0,2)\text{.}$ Note that the interval is centered at $x = 1$ and has radius R = 1.

2. We use the Ratio Test with $a_k = k|x|^k\text{:}$

\begin{align*} \lim_{k \to \infty} \frac{ (k+1)|x|^{k+1} }{ k|x|^k } \amp = |x|\lim_{k \to \infty} \frac{k+1}{k}\\ \amp = |x|\text{.} \end{align*}

So the power series $\sum_{k=1}^{\infty} kx^k$ converges absolutely when $|x| \lt 1$ or when $-1 \lt x \lt 1$ and diverges outside this interval. To completely determine the interval of convergence, we need to check what happens at the endpoints of this interval.

• When $x=-1$ our power series is $\sum_{k=1}^{\infty} (-1)^k k\text{.}$ Since $k \to \infty$ as $k \to \infty\text{,}$ this series diverges by the Divergence Test.

• When $x=1$ our power series is $\sum_{k=1}^{\infty} k$ which again diverges by the Divergence Test.

Therefore, the interval of convergence of the power series $\sum_{k=1}^{\infty} kx^k$ is $(-1,1)\text{.}$ Note that the interval is centered at $x = 0$ and has radius $R = 1\text{.}$

3. We use the Ratio Test with $a_k = \frac{k^2|x+1|^k}{4^k}\text{:}$

\begin{align*} \lim_{k \to \infty} \frac{ \frac{(k+1)^2|x+1|^{k+1}}{4^{k+1}} }{ \frac{k^2|x+1|^k}{4^k} } \amp = \lim_{k \to \infty} \frac{4^k(k+1)^2|x+1|^{k+1}}{4^{k+1}k^2|x+1|^k}\\ \amp = \frac{1}{4}|x+1| \lim_{k \to \infty} \left(\frac{k+1}{k}\right)^2\\ \amp = \frac{1}{4}|x+1|\text{.} \end{align*}

So the power series $\sum_{k=1}^{\infty} \frac{k^2(x+1)^k}{4^k}$ converges absolutely when $\frac{1}{4}|x+1| \lt 1$ or when $-5 \lt x \lt 3$ and diverges outside this interval. To completely determine the interval of convergence, we need to check what happens at the endpoints of this interval.

• When $x=-5$ our power series is $\sum_{k=1}^{\infty} (-1)^k k^2\text{.}$ Since $k^2 \to \infty$ as $k \to \infty\text{,}$ this series diverges by the Divergence Test.

• When $x=3$ our power series is $\sum_{k=1}^{\infty} k^2\text{,}$ which again diverges by the Divergence Test.

Therefore, the interval of convergence of the power series $\sum_{k=1}^{\infty} \frac{k^2(x+1)^k}{4^k}$ is $(-5,3)\text{.}$ Note that the interval is centered at $x = -1$ and has radius $R = 4\text{.}$

4. We use the Ratio Test with $a_k = \frac{|x|^k}{(2k)!}\text{:}$

\begin{align*} \lim_{k \to \infty} \frac{ \frac{|x|^{k+1}}{(2(k+1))!} }{ \frac{|x|^k}{(2k)!} } \amp = \lim_{k \to \infty} |x|\frac{(2k)!}{(2(k+1))!}\\ \amp = |x| \lim_{k \to \infty} \frac{1}{(2k+2)(2k+1)}\\ \amp = 0\text{.} \end{align*}

So the power series $\sum_{k=1}^{\infty} \frac{x^k}{(2k)!}$ converges absolutely on the interval $(-\infty, \infty)\text{.}$ Note that the interval is centered at $x = 0$ and has radius $R = \infty\text{.}$

5. We use the Ratio Test with $a_k = k!|x|^k\text{:}$

\begin{align*} \lim_{k \to \infty} \frac{ (k+1)!|x|^{k+1} }{ k!|x|^k} \amp = \lim_{k \to \infty} |x|(k+1)\\ \amp = \infty \end{align*}

unless $x=0\text{.}$ So the interval of convergence of the power series $\sum_{k=1}^{\infty} \frac{x^k}{k!}$ is $\{0\}\text{.}$ Note that the interval is centered at $x = 0$ and has radius $R = 0\text{.}$

###### Example7.51

Suppose that the power series

\begin{equation*} \sum_{n=1}^\infty C_n (x+2)^n \end{equation*}

converges at $x=2$ and diverges at $x=-8\text{.}$

1. Does the series converge or diverge at $x=-5\text{?}$

2. Does the series converge or diverge at $x=5\text{?}$

3. What are all the possibilities for the radius of convergence of the power series?

1. The series converges at $x=-5.$

2. The series diverges at $x=5.$

3. The radius of converges is at least 4 and at most 6; that is, $4 \leq R \leq 6\text{.}$

Solution

Since the power series is centered at $x=-2\text{,}$ the fact that the series converges at $x=2$ tells us that the radius of convergence is greater than or equal to 4, since 4 is the distance between 2 and -2. Similarly, since the series diverges at $x=-8\text{,}$ this tells us that the radius of convergence is less than or equal to 6. Thus, we have $4 \leq R \leq 6\text{,}$ and we are guaranteed that if an $x$ is less than 4 away from $-2\text{,}$ then the series will converge there. Similarly, if an $x$ is more than 6 away from -2, the series must diverge.

Thus, since -5 is 3 away from -2, and 5 is 7 away from -2, the series will converge at $x=-5$ and diverge at $x=5\text{.}$

### SubsectionSummary

• A power series is a series of the form

\begin{equation*} \sum_{k=0}^{\infty} a_kx^k\text{.} \end{equation*}
• A power series always converges at at least one point. If the power series is centered at $x = a \text{,}$ the power series either converges only at $x = a \text{,}$ or it converges for all $x \in (-\infty, +\infty) \text{,}$ or it converges for all $x$ in a finite interval $(a - R, a + R)$ where $R$ is the radius of convergence. In the latter case, the power series may or may not converge at the endpoints $x = a - R$ and $x = a+R \text{,}$ so these points have to be checked separately.

### SubsectionExercises

Find the radius of convergence of the following series:

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

Consider the power series

\begin{equation*} \sum_{n=1}^\infty C_n(x-1)^n\text{,} \end{equation*}

with radius of convergence $R\text{.}$

1. If the series converges at $x=3$ and diverges at $x=5\text{,}$ what are the possible values for $R\text{?}$

2. If the series converges at $x=4$ and diverges at $x=-2\text{,}$ what are the possible values for $R\text{?}$

3. Is $-3 \lt x \lt 3$ a possible interval of convergence for the series?

4. Is $-2 \leq x \lt 4$ a possible interval of convergence for the series?

In this exercise we will begin with a strange power series and then find its sum. The Fibonacci sequence $\{f_n\}$ is a famous sequence whose first few terms are

\begin{equation*} f_0 = 0, f_1 = 1, f_2 = 1, f_3 = 2, f_4 = 3, f_5 = 5, f_6 = 8, f_7 = 13, \cdots\text{,} \end{equation*}

where each term in the sequence after the first two is the sum of the preceding two terms. That is, $f_0 = 0\text{,}$ $f_1 = 1$ and for $n \geq 2$ we have

\begin{equation*} f_n = f_{n-1} + f_{n-2}\text{.} \end{equation*}

Now consider the power series

\begin{equation*} F(x) = \sum_{k=0}^{\infty} f_kx^k\text{.} \end{equation*}

We will determine the sum of this power series in this exercise.

1. Explain why each of the following is true.

1. $xF(x) = \sum_{k=1}^{\infty} f_{k-1}x^k$

2. $x^2F(x) = \sum_{k=2}^{\infty} f_{k-2}x^k$

2. Show that

\begin{equation*} F(x) - xF(x) - x^2F(x) = x\text{.} \end{equation*}
3. Now use the equation

\begin{equation*} F(x) - xF(x) - x^2F(x) = x \end{equation*}

to find a simple form for $F(x)$ that doesn't involve a sum.

4. Use a computer algebra system or some other method to calculate the first 8 derivatives of $\frac{x}{1-x-x^2}$ evaluated at 0. Why shouldn't the results surprise you?