Finding radius of convergence of power series I have the question to express $\displaystyle f(x)= \frac{2}{4+x} $ as a series and determine when it converges.
This seems to work out pretty easily to $ \sum_{k=0}^\infty  2(-x-3)^k $ , and this seems to work for values $ -4<x<-2 $
However, the answer in the book is $\displaystyle \sum_{k=0}^\infty  (-1)^k\frac{1}{2^{2k+1}}x^k $, with $ -4<x<4 $.
I keep going over the proof that $ \sum_{k=0}^\infty ar^k $ converges to $\displaystyle \frac{a}{1-r} $ if $ |r| < 1 $ but I can't figure out what I'm missing.
 A: Hint: $$\displaystyle \frac{2}{4+x} = \frac{1}{2}\cdot \frac{1}{1+\frac{x}{4}}$$
I presume from your lectures you know something about the series $\frac{1}{1+x}$
If not, you can get it from the result you've written about geometric series. Try setting $\displaystyle a=\frac{1}{2}$ and $\displaystyle  r=-\frac{x}{4}$ 
A: Yes. You have stumbled upon the fact that Taylor expansions of an analylic function are local. Both your answer and the book's answer are correct. You have provided the Taylor expansion of the rational function $2(x+4)^{-1}$ around $3$ (upon substituting $x$ for $-x$ in your above expression) whereas the book gave the expansion around $0.$ In fact, for every $x_0\in \mathbb{R}$ with the exception of the case where $x_0 = 4,$ there exists an interval $I =(x_0-a,x_0+a)$ and a power series $p_{x_0}(x) = \sum a_i(x-x_0)^i$ which converges on $I$ such that $f(x) = p_{x_0}(x)$ for all $x$ in $I.$  
It might be a helpful exercise to find some Taylor expansions for $f$ around other points. For example, can you find an interval $I$ around $-1$ and a power series which converges on that $I$ such that the evaluation of that power series at any point $x$ of $I$ is equal to $f(x)$ ?        
A: Note that: $$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n \, ,\,\,\,\,\,\text{provided}\,\,\, |x|\lt 1.$$ So using Kuch's hint $$\frac{2}{4+x}=\frac{1}{2}.\frac{1}{1-(-\frac{x}{4})}.$$
And  $$\frac{1}{1-(-\frac{x}{4})}=\sum_{n=0}^{\infty}\left(\frac{-x}{4}\right)^n\,, \,\,\,\,\,\text{provided} \,\,\,\,\,\left|\frac{-x}{4}\right|\lt 1.$$
Combining everything thus far should clear things up.
