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
(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}}

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<x+7<1\quad\Longrightarrow\quad-8<x<-6.
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.


About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Analysis, Calculus, Infinite Series, Power Series and tagged . Bookmark the permalink.

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