<s>[INST] Compute Taylor series $\frac{1}{x^2+4x+3}$ at $x = 2$ I was trying to solve a textbook exercise stated in the following: 
Use completing the square and the geometric series to get the Taylor expansion about $x=2$ of $
\frac{1}{x^2+4x+3}$
My early attempt was 
$\frac{1}{x^2+4x+3} = -1 (\frac{1}{1-(x+2)^2})$, even though the expression inside the parenthesis is of the form of geometric series $\frac{1}{1 - x}$. I realized that $x=2$ is not in the convergence domain of geometric series i.e. $|(x+2)^2| < 1$. So I should be wrong to proceed in this direction.  
Could you please provide me some other directions to work with?
 [/INST] Using partial fraction decomposition we get that
$$\frac{1}{x^2+4x+3} = \frac{1}{(x+3)(x+1)} = \frac{1}{2}\left(\frac{1}{x+1} - \frac{1}{x+3}\right) = \frac{1}{2}\left(\frac{1}{3}\frac{1}{1+\frac{x-2}{3}} - \frac{1}{5}\frac{1}{1+\frac{x-2}{5}} \right)$$
where we can now use geometric series to get
$$\frac{1}{2}\sum_{n=0}^\infty\left(\frac{1}{3^{n+1}}-\frac{1}{5^{n+1}}\right)(-1)^n(x-2)^n$$
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