A power series is a series of the form \begin{equation}\label{eq:powerseries}\sum_{n=0}^\infty c_n(x-a)^n=c_0+c_1(x-a)+c_2(x-a)^2+\cdots\end{equation} The power series in \eqref{eq:powerseries} is particularly called a *power series in $x-a$*, a *power series centered at $a$*, or a *power series about $a$*. The number $a$ is called the *center of the power series*. The set of the values of $x$ for which the series converges is called the *interval of convergence* and the radius of the interval of convergence is called the *radius of convergence*. Since a power series in general does not converge everywhere, it is important to find the interval of convergence or the radius of convergence (if you find one, you consequently find the other). In the study of the convergence of power series, we mostly use the ratio test but the root test may come in handy sometimes.

*Example*. Find the interval and the radius of convergence for each of the following series.

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

*Solution*.

- \begin{align*}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\frac{\frac{|x|^{n+1}}{(n+1)!}}{\frac{|x|^n}{n!}}\\&=\lim_{n\to\infty}\frac{|x|}{n+1}\\&=0<1\end{align*} for all $x$. This means that the interval of convergence is $(-\infty,\infty)$ and the radius of convergence is $\infty$.
- \begin{align*}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\frac{(n+1)!|x|^{n+1}}{n!|x|^n}\\&=\lim_{n\to\infty}(n+1)|x|\\&=\infty\end{align*} unless $x=0$. The interval of convergence is $\{0\}$ and the radius of convergence is 0.
- \begin{align*}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\frac{\frac{|x-3|^{n+1}}{n+1}}{\frac{|x-3|^n}{n}}\\&=\lim_{n\to\infty}\frac{n}{n+1}|x-3|\\&=|x-3|\end{align*} In order for the series to converge, we require that $|x-3|<1$ or $-1<x-3<1$ i.e. $2<x<4$. For the last step, we must check if the series converges at the end points $x=2$ and $x=4$. If $x=2$, the series becomes $\sum_{n=1}^\infty\frac{(-1)^n}{n}$. This is an alternating harmonic series and we learned here that it converges. If $x=4$, the series becomes the harmonic series $\sum_{n=1}^\infty\frac{1}{n}$ which diverges. Therefore, the interval of convergence is $[2,4)$. Since the center of the power series is 3, the radius is 1.

**Representing Functions as Power Series**

As we will study later, functions that are infinitely many times differentiable (such functions are simply called *smooth*) in a open interval containing $a$ can be represented a power series centered at $a$ called a *Taylor series*. But even without knowing Taylor series expansion of a function, using the geometric series \begin{equation}\label{eq:geomseries}\frac{1}{1-x}=1+x+x^2+x^3+\cdots=\sum_{n=0}^\infty x^n\ \mbox{if}\ |x|<1\end{equation} we can represent many functions as power series.

*Example*. Express $\frac{1}{1+x^2}$ as a power series and find the interval of convergence.

*Solution*. \begin{align*}\frac{1}{1+x^2}&=\frac{1}{1-(-x^2)}\\&=\sum_{n=0}^\infty(-x^2)^n\\&=\sum_{n=0}^\infty(-1)^nx^{2n}\end{align*} provided $|x|^2<1$. Hence, the interval of convergence is $|x|<1$ or $(-1,1)$.

*Example*. Find a power series representation of $\frac{x^3}{x+2}$.

*Solution*. First we find a power series representation of $\frac{1}{x+2}$. \begin{align*}\frac{1}{x+2}&=\frac{1}{2}\frac{1}{1+\frac{x}{2}}\\&=\frac{1}{2}\frac{1}{1-\left(-\frac{x}{2}\right)}\\&=\frac{1}{2}\sum_{n=0}^\infty(-1)^n\frac{x^n}{2^n}\\&=\sum_{n=0}^\infty(-1)^n\frac{x^n}{2^{n+1}}\end{align*} provided $\left|\frac{x}{2}\right|<1$ or $|x|<2$. Therefore, $\frac{x^3}{x+2}=\sum_{n=0}^\infty(-1)^n\frac{x^{n+3}}{2^{n+1}}$ for $-2<x<2$.

Note: If one replaces $n$ by $n-3$, then the series can be rewritten equivalently as $\sum_{n=3}^\infty(-1)^{n-1}\frac{x^n}{2^{n-2}}$.

**Differentiation and Integration of Power Series**

It turns out that one can differentiate and integrate a power series term by term as he would’ve done with polynomials.

*Theorem*. If the power series $\sum_{n=0}^\infty c_n(x-a)^n$ has radius of convergence $R>0$ then the function $f$ defined by $$f(x)=\sum_{n=0}^\infty c_n(x-a)^n$$ is differentiable on the interval of convergence $(a-R,a+R)$ and $$f'(x)=\sum_{n=1}^\infty nc_n(x-a)^{n-1}$$ and $$\int f(x)dx=C+\sum_{n=0}^\infty\frac{c_n}{n+1}(x-a)^{n+1}$$ where $C$ is a constant. The two power series $f'(x)$ and $\int f(x)dx$ both have the same radius of convergence $R$.

This theorem along with \eqref{eq:geomseries} allows us to find the power series representations of a broader class of functions as we will see in the following examples.

*Example*. Express $\frac{1}{(1-x)^2}$ as a power series. What is the radius of convergence?

*Solution*. \begin{align*}\frac{1}{(1-x)^2}&=\frac{d}{dx}\frac{1}{1-x}\\&=\frac{d}{dx}\sum_{n=0}^\infty x^n\ (|x|<1)\\&=\sum_{n=1}^\infty nx^{n-1}\\&=\sum_{n=0}^\infty(n+1)x^n\end{align*} The radius of convergence is 1.

*Example*. Find a power series representation for $\ln (1-x)$ and its radius of convergence.

*Solution*. \begin{align*}\ln(1-x)&=-\int\frac{dx}{1-x}\\&=-\int\sum_{n=0}^\infty x^ndx\ (|x|<1)\\&=-\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}+C\\&=-\sum_{n=1}^\infty\frac{x^n}{n}+C\end{align*} where $C$ is a constant. When $x=0$, $\ln(1-x)=0$ so we find $C=0$. Therefore, $$\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}$$

*Example*. Find a power series representation for $f(x)=\tan^{-1}x$.

*Solution*. \begin{align*}\tan^{-1} x&=\int\frac{1}{1+x^2}dx\\&=\int\sum_{n=0}^\infty (-1)^nx^{2n}\ (|x|<1)\\&=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}+C\end{align*} where $C$ is a constant. Since $\tan^{-1}0=0$, we find $C=0$. Hence, $$\tan^{-1}x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$$

*Example*. Find a power series representation of $\ln\left(\frac{1+x}{1-x}\right)$.

*Solution*. First $\ln\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln(1-x)$. \begin{align*}\ln(1+x)&=\int\frac{dx}{1+x}\\&=\int\sum_{n=0}^\infty (-1)^nx^ndx\ (|x|<1)\\&=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}+C\end{align*} Using $\ln(1+x)=0$ for $x=0$, we find $C=0$ and so $\ln(1+x)=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}$. Recall that we found a power series representation $\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}$ in an example above. Hence, \begin{align*}\ln\left(\frac{1+x}{1-x}\right)&=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}+\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}\\&=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n}+\sum_{n=1}^\infty\frac{x^n}{n}\\&=\sum_{n=1}^\infty[(-1)^{n-1}+1]\frac{x^n}{n}\end{align*} Note $$(-1)^{n-1}+1=\left\{\begin{array}{ccc} 2 & \mbox{if} & n\ \mbox{is odd}\\0 & \mbox{if} & n\ \mbox{is even}\end{array}\right.$$ Therefore, $$\ln\left(\frac{1+x}{1-x}\right)=2\sum_{n=1}^\infty\frac{x^{2n-1}}{2n-1}=2\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}$$

*Remark*. Since we already know that $\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}$ and $\ln(1+x)=\ln(1-(-x))$, we can find a power series representation of $\ln(1+x)$ by replacing $x$ by $-x$ in the power series representation of $\ln(1-x)$: \begin{align*}\ln(1+x)&=-\sum_{n=0}^\infty\frac{(-x)^{n+1}}{n+1}\\&=-\sum_{n=0}^\infty\frac{(-1)^{n+1}x^{n+1}}{n+1}\\&=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}\end{align*}