So far, our study of series has examined the question of “Is the sum of these infinite terms finite?,” i.e., “Does the series converge?” We now approach series from a different perspective: as a function. Given a value of $x\text{,}$ we evaluate $f(x)$ by finding the sum of a particular series that depends on $x$ (assuming the series converges). We start this new approach to series with a definition.

##### Definition8.6.1Power Series

Let $\{a_n\}$ be a sequence, let $x$ be a variable, and let $c$ be a real number.

1. The power series in $x$ is the series \begin{equation*} \infser[0] a_nx^n = a_0+a_1x+a_2x^2+a_3x^3+\ldots \end{equation*}

2. The power series in $x$ centered at $c$ is the series \begin{equation*} \infser[0] a_n(x-c)^n = a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+\ldots \end{equation*}

##### Example8.6.2Examples of power series

Write out the first five terms of the following power series:

$\ds 1.\ \infser[0] x^n \qquad\qquad 2.\ \infser (-1)^{n+1}\frac{(x+1)^n}n\qquad\qquad 3.\ \infser[0] (-1)^{n+1} \frac{(x-\pi)^{2n}}{(2n)!}\text{.}$

Solution

We introduced power series as a type of function, where a value of $x$ is given and the sum of a series is returned. Of course, not every series converges. For instance, in part 1 of Example 8.6.2, we recognized the series $\ds \infser[0] x^n$ as a geometric series in $x\text{.}$ Theorem 8.2.5 states that this series converges only when $\abs{x}\lt 1\text{.}$

This raises the question: “For what values of $x$ will a given power series converge?,” which leads us to a theorem and definition.

The value of $R$ is important when understanding a power series, hence it is given a name in the following definition. Also, note that part 2 of Theorem 8.6.3 makes a statement about the interval $(c-R,c+R)\text{,}$ but the not the endpoints of that interval. A series may/may not converge at these endpoints.

##### Definition8.6.4Radius and Interval of Convergence
1. The number $R$ given in Theorem 8.6.3 is the radius of convergence of a given series. When a series converges for only $x=c\text{,}$ we say the radius of convergence is 0, i.e., $R=0\text{.}$ When a series converges for all $x\text{,}$ we say the series has an infinite radius of convergence, i.e., $R=\infty\text{.}$

2. The interval of convergence is the set of all values of $x$ for which the series converges.

To find the values of $x$ for which a given series converges, we will use the convergence tests we studied previously (especially the Ratio Test). However, the tests all required that the terms of a series be positive. The following theorem gives us a work–around to this problem.

Theorem 8.6.5 allows us to find the radius of convergence $R$ of a series by applying the Ratio Test (or any applicable test) to the absolute value of the terms of the series. We practice this in the following example.

##### Example8.6.6Determining the radius and interval of convergence.

Find the radius and interval of convergence for each of the following series:

1. $\ds \infser[0] \frac{x^n}{n!}$

2. $\ds \infser (-1)^{n+1}\frac{x^n}{n}$

3. $\ds \infser[0] 2^n(x-3)^n$

4. $\ds \infser[0] \abs{\frac{x^n}{n!}}$

Solution

We can use a power series to define a function: \begin{equation*} f(x) = \infser[0] a_nx^n \end{equation*} where the domain of $f$ is a subset of the interval of convergence of the power series. One can apply calculus techniques to such functions; in particular, we can find derivatives and antiderivatives.

A few notes about Theorem 8.6.7:

1. The theorem states that differentiation and integration do not change the radius of convergence. It does not state anything about the interval of convergence. They are not always the same.

2. Notice how the summation for $\fp(x)$ starts with $n=1\text{.}$ This is because the constant term $a_0$ of $f(x)$ becomes $0$ through differentiation.

3. Differentiation and integration are simply calculated term–by–term using the Power Rules.

##### Example8.6.8Derivatives and indefinite integrals of power series

Let $\ds f(x) = \infser[0] x^n\text{.}$ Find $\fp(x)$ and $\ds F(x) =\int f(x)\ dx\text{,}$ along with their respective intervals of convergence.

Solution

The previous example showed how to take the derivative and indefinite integral of a power series without motivation for why we care about such operations. We may care for the sheer mathematical enjoyment “that we can”, which is motivation enough for many. However, we would be remiss to not recognize that we can learn a great deal from taking derivatives and indefinite integrals.

Recall that $\ds f(x) = \infser[0] x^n$ in Example 8.6.8 is a geometric series. According to Theorem 8.2.5, this series converges to $1/(1-x)$ when $\abs{x}\lt 1\text{.}$ Thus we can say \begin{equation*} f(x) = \infser[0] x^n = \frac 1{1-x}, \text{ on } (-1,1). \end{equation*}

Integrating the power series, (as done in Example 8.6.8,) we find $$F(x) = C_1+\infser[0] \frac{x^{n+1}}{n+1}, \label{eq_ps3a}\tag{8.6.1}$$ while integrating the function $f(x) = 1/(1-x)$ gives $$F(x) = -\ln\abs{1-x} + C_2. \label{eq_ps3b}\tag{8.6.2}$$

Equating Equations (8.6.1) and (8.6.2), we have \begin{equation*} F(x) = C_1+\infser[0] \frac{x^{n+1}}{n+1} = -\ln\abs{1-x} + C_2. \end{equation*}

Letting $x=0\text{,}$ we have $F(0) = C_1 = C_2\text{.}$ This implies that we can drop the constants and conclude \begin{equation*} \infser[0] \frac{x^{n+1}}{n+1} = -\ln\abs{1-x}. \end{equation*}

We established in Example 8.6.8 that the series on the left converges at $x=-1\text{;}$ substituting $x=-1$ on both sides of the above equality gives \begin{equation*} -1+\frac12-\frac13+\frac14-\frac15+\cdots = -\ln(2) . \end{equation*}

On the left we have the opposite of the Alternating Harmonic Series; on the right, we have $-\ln(2)\text{.}$ We conclude that \begin{equation*} 1-\frac12+\frac13-\frac14+\cdots = \ln(2) . \end{equation*}

Important: We stated in Key Idea 8.2.17 (in Section 8.2) that the Alternating Harmonic Series converges to $\ln(2)\text{,}$ and referred to this fact again in Example 8.5.4 of Section 8.5. However, we never gave an argument for why this was the case. The work above finally shows how we conclude that the Alternating Harmonic Series converges to $\ln(2)\text{.}$

We use this type of analysis in the next example.

##### Example8.6.9Analyzing power series functions

Let $\ds f(x) = \infser[0] \frac{x^n}{n!}\text{.}$ Find $\ds \fp(x)$ and $\ds \int f(x)\ dx\text{,}$ and use these to analyze the behavior of $f(x)\text{.}$

Solution

Example 8.6.9 and the work following Example 8.6.8 established relationships between a power series function and “regular” functions that we have dealt with in the past. In general, given a power series function, it is difficult (if not impossible) to express the function in terms of elementary functions. We chose examples where things worked out nicely.

In this section's last example, we show how to solve a simple differential equation with a power series.

##### Example8.6.10Solving a differential equation with a power series.

Give the first 4 terms of the power series solution to $y' = 2y\text{,}$ where $y(0) = 1\text{.}$

Solution

Our last example illustrates that it can be difficult to recognize an elementary function by its power series expansion. It is far easier to start with a known function, expressed in terms of elementary functions, and represent it as a power series function. One may wonder why we would bother doing so, as the latter function probably seems more complicated. In the next two sections, we show both how to do this and why such a process can be beneficial.

# Subsection8.6.1Exercises

Terms and Concepts

In the following exercises, write out the sum of the first 5 terms of the given power series.