## Power Series

1. Definitions. Let ${x}$ be a variable. a series of the form
(1)$\displaystyle \sum^{\infty}_{n=0}a_{n}x^{n}=a_{0}+a_{1}x+a_{2}x^{2}+\dots+a_{n}x^{n}+\dots$
is called a “Power Series about ${x=0}$.

A series of the form
(2)$\displaystyle \sum^{\infty}_{n=0}b_{n}(x-a)^{n}=b_{0}+b_{1}(x-a)+b_{2}(x-a)^{2}+\dots+b_{n}(x-a)^{n}+\dots$
is called a “Power Series about ${x=a}$.

Example 1. Consider the series
(3)$\displaystyle f(x)=\sum^{\infty}_{n=0}\frac{n+1}{2^{n}}x^{n}.$
For what values of ${x}$ does this series converge? When will the series diverge?

Solution. Using the absolute ratio test, we see
(4)\displaystyle \begin{aligned} \lim_{n\rightarrow\infty}\left|\frac{\displaystyle\frac{n+2}{2^{n+1}}x^{n+1}}{\displaystyle\frac{n+1}{2^{n}}x^{n}}\right| &=\lim_{n\rightarrow\infty}\frac{n+2}{n+1}\cdot\frac{2^{n}}{2^{n+1}}\cdot|x|\\ &=(1)\cdot\left(\frac{1}{2}\right)\cdot|x|\\ &=\frac{1}{2}|x| \end{aligned}
For convergence, we need
(5)$\displaystyle \frac{1}{2}|x|<1\quad\Longrightarrow\quad|x|<2.$
So divergence would be when
(6)$\displaystyle |x|>2.$
We need to check the ${|x|=2}$ case. Observe, for ${x=2}$, we have
(7)$\displaystyle f(2)=\sum^{\infty}_{n=0}\frac{n+1}{2^{n}}2^{n}=\sum^{\infty}_{n=0}n+1$
which diverges. And at ${x=-2}$ we have
(8)$\displaystyle f(-2)=\sum^{\infty}_{n=0}\frac{n+1}{2^{n}}(-2)^{n}=\sum^{\infty}_{n=0}(-1)^{n}(n+1)$
which still diverges!

2. Theorem. Let ${c\not=0}$ and
(9)$\displaystyle f(x)=\sum^{\infty}_{n=0}a_{n}x^{n}$
converge at ${x=c}$. Then ${f(x)}$ converges absolutely for ${|x|<|c|}$.

Proof: Let ${x}$ be any value such that ${|x|<|c|}$. Since ${f(c)}$ converges, there exists an ${M}$ such that
(10)$\displaystyle |a_{n}c^{n}|\leq M\quad\mbox{for any }n.$
Well, we see
(11)$\displaystyle |x/c|\leq 1$
so
(12)$\displaystyle a_{n}x^{n}=a_{n}c^{n}\left(\frac{x}{c}\right)^{n}$
and moreover
(13)$\displaystyle |a_{n}x^{n}| = |a_{n}c^{n}|\cdot\left|\frac{x}{c}\right|^{n}\leq M\left|\frac{x}{c}\right|^{n}$
But observe the series
(14)$\displaystyle g(x)=\sum^{\infty}_{n=0}M\cdot\left|\frac{x}{c}\right|^{n}$
is a geometric series which converges since ${|x/c|\leq1}$.

Therefre the series ${\sum|a_{n}x^{n}|}$ converges by comparison, and the absolute convergence test tells us ${\sum a_{n}x^{n}}$ converges. Therefore ${\sum a_{n}x^{n}}$ is absolutely convergent.

Example 2. Find the values of ${x}$ for which
(15)$\displaystyle \sum^{\infty}_{n=1}\frac{(x+7)^{n}}{\sqrt{n}}$
converge.

Solution: Using the absolute ratio test, we find
(16)\displaystyle \begin{aligned} \lim_{n\rightarrow\infty}\left|\frac{\left(\displaystyle\frac{(x+7)^{n+1}}{\sqrt{n+1}}\right)}{\left(\displaystyle\frac{(x+7)^{n}}{\sqrt{n}}\right)}\right| &=\lim_{n\rightarrow\infty}\frac{\sqrt{n}}{\sqrt{n+1}}|x+7|\\ &=|x+7|. \end{aligned}
We get convergence for ${|x+7|<1}$. So
(17)$\displaystyle -1
We have to check the boundary cases.

When ${x=-6}$, we have
(18)$\displaystyle \sum^{\infty}\frac{(7-6)^{n}}{\sqrt{n}}=\sum^{\infty}_{n=1}\frac{1}{\sqrt{n}}$
which diverges by the integral test (or the comparison test with the Harmonic series). For ${x=-8}$ we have our series become
(19)$\displaystyle \sum^{\infty}_{n=1}\frac{(-1)^{n}}{\sqrt{n}}$
which converges by the alternating series test.

Exercise 1. Find the values of $x$ for which
$\displaystyle\sum^{\infty}_{n=0}\frac{(x-5)^{n}}{\sqrt{2n+3}}$ converges.