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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.

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See math.stackexchange.com/q/47151/11619 for a very similar problem. –  Jyrki Lahtonen Jun 24 '11 at 5:45
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3 Answers

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 $\displaystyle r=-\frac{x}{4}$

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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 $f(x)$ ?

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"Both your answer and the book's answer are correct." That would depend on the exact statement and context of the question. If it's clear from context that the only acceptable answer is a series in powers of $x$, then Gerber's answer is not correct. Aside from that, I fully endorse what you say and hope Gerber takes encouragement from it. –  Gerry Myerson Jun 24 '11 at 6:13
This is very interesting. If I go down this road will it help explain why the radius of convergence is smaller for my answer than the answer in powers of x even though the answers seem equivalent? Does the series in powers of x always have the widest radius of convergence compared to x+/-c where c is some constant? –  Gerber Jun 24 '11 at 6:58
@Gerber: A course in complex analysis will give you a nice answer. It will be along the following lines: You see that something bad happens to your function at $x=-4$. Your series is centered at $-3$, the book answer at $0$, so your series has a smaller radius of convergence, because $0$ is further away from the bad point. You need to be careful about the complex part, though. For example, the function $1/(1+x^2)$ has a bad point at $x=i$, which is at distance 1 from the origin. Therefore the radius of convergence of its Maclaurin series cannot exceed 1. –  Jyrki Lahtonen Jun 24 '11 at 7:54
I'm looking forward to complex analysis. I find that $\displaystyle \sum_{k=0}^\infty \frac{1}{4}(\frac{1}{2}-\frac{x}{8})^k $ is also equivalent with radius 8. Seems like I can expand the radius indefinitely, but always with the lower bound -4? –  Gerber Jun 25 '11 at 3:41
@Gerber: Correct. $x=-4$ is the only bad point, so if your series is centered at $x_0>0$, then the radius of convergence will be $R=x_0+4$, and you can make this as large as you wish. –  Jyrki Lahtonen Jun 30 '11 at 9:46
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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.

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