<s>[INST] Taylor expansion about a point I need help with the following calculus problem:
Use completing the square and the geometric series to get the Taylor expansion about ${x=2}$  of ${\frac{1}{x^{2}+4x+3}}$
So far I have the following:
 By completing the square. It can be shown that
${\frac{1}{x^{2}+4x+3} = \frac{1}{(x+2)^{2}-1}}$
We want to use the geometric series, so we recognize that 
${\frac{1}{(x+2)^{2}-1} = -1\times\frac{1}{1-(x+2)^{2}}}$ 
Using what we know about the geometric series, it is seen that
${\frac{1}{1-(x+2)^2} = \sum\limits_{k=0}^\infty (x+2)^{2k}}$ and our Taylor series about ${x = -2}$ is ${-\sum\limits_{0}^{\infty}(x+2)^{2k}}$
I am stuck, however on figuring out how to use this expansion about ${x=-2}$ to find the expansion about ${x=2}$. How is this done, without calculating the derivative at ${x=2}$ of the original function and trying to find a pattern for the summation?
 [/INST] $$\frac{1}{x^2+4x+3}=\frac{1}{(x+1)(x+3)}=\frac{1}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}\right)=\frac{1}{2}\left(\frac{1}{(x-2)+3}-\frac{1}{(x-2)+5}\right)$$
Now just use:
$$\frac{1}{z+3}=\frac{\frac{1}{3}}{1+\frac{z}{3}}=\sum_{n\geq 0}\frac{(-1)^n}{3^{n+1}}z^n $$
to get:
$$\frac{1}{x^2+4x+3}=\frac{1}{2}\sum_{n\geq 0}\left(\frac{1}{3^{n+1}}-\frac{1}{5^{n+1}}\right)(-1)^n (x-2)^n.$$
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