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## Power Series

A power series is a series of the form $\sum_{n=0}^{\infty} a_n x^n=a_0+a_1x+a_2x^2+a_3x^3+\cdots.$ A power series about $$c$$ (or, centered at $$c$$) is a series of the form $\sum_{n=0}^{\infty} a_n (x-c)^n=a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+\cdots.$

Example.

1. $$\displaystyle\sum_{n=0}^{\infty} x^n=1+x+x^2+x^3+\cdots$$ is a power series about $$0$$.
2. $$\displaystyle\sum_{n=0}^{\infty} \frac{(x-2)^n}{n!}=1+\frac{(x-2)}{1!}+\frac{(x-2)^2}{2!}+\frac{(x-2)^3}{3!}+\cdots$$ is a power series about $$2$$.

Theorem. There are three possibilities regarding convergence of a power series $$\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$$:

1. $$\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$$ converges only for $$x=c$$.
2. $$\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$$ converges for all values of $$x$$.
3. There is a positive real number $$R$$ such that $$\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$$ converges if $$|x-c| < R$$ and diverges if $$|x-c| > R$$.

Example. The power series $$\displaystyle\sum_{n=0}^{\infty} x^n=1+x+x^2+x^3+\cdots$$ is a geometric series with the common ratio $$x$$. Therefore it converges if $$|x| < 1$$ and diverges if $$|x|\geq 1$$. So in this case, $$R=1$$.

Definition. If $$\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$$ converges for $$|x-c| < R$$ and diverges for $$|x-c| > R$$, then $$R$$ is called the radius of convergence of $$\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$$. We define $$R$$ to be $$0$$ or $$\infty$$ if $$\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$$ converges only for $$x=c$$ and for all values of $$x$$ respectively. The interval of convergence of $$\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$$ is the interval that contains all the values of $$x$$ for which $$\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$$ converges.

Note that the interval of convergence contains $$(c-R,c+R)$$.

Example. Find the radius of convergence and the interval of convergence of the power series $\displaystyle\sum_{n=1}^{\infty} \frac{(2x-1)^n}{3^n\sqrt{n}}.$
Solution. Here $a_n=\frac{(2x-1)^n}{3^n\sqrt{n}}=\frac{2^n \left(x-\frac{1}{2}\right)^n}{3^n\sqrt{n}}=\left(\frac{2}{3}\right)^n \frac{\left(x-\frac{1}{2}\right)^n}{\sqrt{n}}.$ Then $a_{n+1}=\left(\frac{2}{3}\right)^{n+1} \frac{\left(x-\frac{1}{2}\right)^{n+1}}{\sqrt{n+1}}$ and $\frac{a_{n+1}}{a_n}=\left(\frac{2}{3}\right)^{n+1} \frac{\left(x-\frac{1}{2}\right)^{n+1}}{\sqrt{n+1}} \left(\frac{3}{2}\right)^n \frac{\sqrt{n}}{\left(x-\frac{1}{2}\right)^n} =\frac{2}{3} \left(x-\frac{1}{2}\right) \sqrt{\frac{n}{n+1}}.$ $\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert =\lim_{n\to \infty} \left\vert \frac{2}{3} \left(x-\frac{1}{2}\right) \sqrt{\frac{n}{n+1}} \right\vert =\lim_{n\to \infty} \frac{2}{3} \left\vert x-\frac{1}{2} \right\vert \sqrt{\frac{1}{1+\frac{1}{n}}} =\frac{2}{3} \left\vert x-\frac{1}{2} \right\vert.$ By the Ratio Test, the given power series is convergent if $$\frac{2}{3} \left\vert x-\frac{1}{2} \right\vert < 1$$, i.e., $$\left\vert x-\frac{1}{2} \right\vert< \frac{3}{2}$$ and divergent if $$\frac{2}{3} \left\vert x-\frac{1}{2} \right\vert > 1$$, i.e., $$\left\vert x-\frac{1}{2} \right\vert > \frac{3}{2}$$. Thus the radius of convergence is $$R=\frac{3}{2}$$.

Since $$\left\vert x-\frac{1}{2} \right\vert < \frac{3}{2} \implies \frac{1}{2}-\frac{3}{2} < x < \frac{1}{2}+\frac{3}{2} \implies -1 < x < 2$$, the interval of convergence contains $$(-1,2)$$. Now we need to determine the convergence for $$x=-1,2$$ (the endpoints).

At $$x=2$$, the power series becomes $$\displaystyle\sum_{n=1}^{\infty} \frac{(2\cdot 2-1)^n}{3^n\sqrt{n}}= \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$$ which is divergent by the $$p$$-series Test with $$p=\frac{1}{2}\leq 1$$.

At $$x=-1$$, the power series becomes $$\displaystyle\sum_{n=1}^{\infty} \frac{(2(-1)-1)^n}{3^n\sqrt{n}}= \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$$. Let $$a_n=\frac{1}{\sqrt{n}}$$. For all integers $$n\geq 1$$, $$a_n=\frac{1}{\sqrt{n}} > 0$$ and $$a_{n+1}=\frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}}=a_n$$. Also $$\displaystyle\lim_{n\to \infty} a_n =\lim_{n\to \infty}\frac{1}{\sqrt{n}}=0$$. Thus $$\{a_n\}=\{\frac{1}{\sqrt{n}}\}$$ is a positive and decreasing sequence that converges to $$0$$. By the Alternating Series Test, $$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$$ is convergent.

Since the power series is convergent at $$-1$$ and divergent at $$2$$, the interval of convergence is $$[-1,2)$$.

Definition. Let $$I$$ be the interval of convergence of $$\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$$. The sum function $$f$$ of the series is a function with domain $$I$$ that is defined as $f(x)=\sum_{n=0}^{\infty} a_n (x-c)^n,$ for all $$x$$ in $$I$$.

Example. The power series $$\displaystyle\sum_{n=0}^{\infty} x^n=1+x+x^2+x^3+\cdots$$ converges to $$\frac{1}{1-x}$$ if $$|x| < 1$$ and diverges if $$|x|\geq 1$$. So the sum function is $$f(x)=\frac{1}{1-x}$$ for all $$x$$ in $$I=(-1,1)$$ and we write $\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n=1+x+x^2+x^3+\cdots,\; |x| < 1.$

The above example helps us write some functions as power series as illustrated by the following example.

Example. Find a power series representation of $$\displaystyle\frac{2x}{x^2+9}$$ and its interval of convergence.
Solution. \begin{align*} \frac{2x}{x^2+9}=\frac{2x}{9\left( 1+\frac{x^2}{9} \right)}=\frac{2x}{9} \frac{1}{\left( 1-\left(-\frac{x^2}{9}\right) \right)} &=\frac{2x}{9} \sum_{n=0}^{\infty} \left(-\frac{x^2}{9}\right)^n,\; \left\vert -\frac{x^2}{9}\right\vert <1\\ &=\frac{2x}{9} \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{9^n},\; |x^2| <9\\ &=\sum_{n=0}^{\infty} \frac{(-1)^n2x^{2n+1}}{9^{n+1}},\; |x| <3. \end{align*} Since the power series converges if and only if $$|x| < 3$$, the interval of convergence is $$(-3,3)$$.

Theorem. Suppose that the power series $$f(x)=\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$$ has radius of convergence $$R$$. Then the power series is differentiable and integrable on $$(c-R,c+R)$$. Moreover, the power series can be differentiated and integrated term by term: \begin{align*} f'(x) &=\displaystyle\sum_{n=1}^{\infty} na_n (x-c)^{n-1}=a_1+2a_2(x-c)+3a_3(x-c)^2+\cdots.\\ \int f(x)\;dx &= C+\displaystyle\sum_{n=0}^{\infty} a_n \frac{(x-c)^{n+1}}{n+1}=C+a_0(x-c)+a_1\frac{(x-c)^2}{2}+a_2\frac{(x-c)^3}{3}+\cdots. \end{align*}

The above can alternatively be written as follows: \begin{align*} \frac{d}{dx} \left[ \sum_{n=0}^{\infty} a_n (x-c)^n \right] &=\sum_{n=0}^{\infty} \left[ \frac{d}{dx} a_n (x-c)^n \right].\\ \int \left[ \sum_{n=0}^{\infty} a_n (x-c)^n \right]dx &= \sum_{n=0}^{\infty} \left[ \int a_n (x-c)^n \;dx \right]. \end{align*}

The preceding theorem can be used to find power series representations of some functions.

Example. Find a power series representation of each of the following functions: (a) $$\ln(1+x)$$, (b) $$\frac{1}{(1-x)^2}$$.

Solution. (a) Since $$\displaystyle\int \frac{1}{1+x}\;dx=\ln(1+x)$$ and $\frac{1}{1+x}=\frac{1}{1-(-x)}=\sum_{n=0}^{\infty} (-x)^n=\sum_{n=0}^{\infty} (-1)^nx^n,\; |x| < 1,$ we have \begin{align*} \ln(1+x) &=\int\frac{1}{1+x} \;dx \\ &= \int \left[ \sum_{n=0}^{\infty} (-1)^nx^n \right] \;dx\\ &= \sum_{n=0}^{\infty} \left[ \int(-1)^nx^n \;dx\right] \\ &= C+\sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}. \end{align*} To find $$C$$, we plug $$x=0$$: $\ln(1)=C+\sum_{n=0}^{\infty} (-1)^n \frac{0^{n+1}}{n+1} \implies C=0.$ Thus $\ln(1+x)=\sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots,\; |x| < 1.$

(b) Since $$\displaystyle\frac{d}{dx} \left(\frac{1}{1-x} \right)=\frac{1}{(1-x)^2}$$ and $\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n,\; |x|<1,$ we have \begin{align*} \frac{1}{(1-x)^2} &= \frac{d}{dx} \left(\frac{1}{1-x} \right)\\ &= \frac{d}{dx} \left[ \sum_{n=0}^{\infty} x^n \right]\\ &= \sum_{n=0}^{\infty} \left[ \frac{d}{dx} (x^n) \right]\\ &= \sum_{n=1}^{\infty} nx^{n-1},\; |x|<1. \end{align*}

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