<s>[INST] Find the Taylor expansion of $\frac{1}{x^2+2x-3}$ around $x=-1$. Find the Taylor expansion of $\frac{1}{x^2+2x-3}$ around $x=-1$. What is its radius of convergence?
So I write the fraction as $\frac{1}{(x-1)(x+3)}$ and what should I do now?
 [/INST] You may write
$$
x^2+2x-3=-4+(x+1)^2
$$ then
$$
\frac{1}{x^2+2x-3}=\frac{1}{-4+(x+1)^2}=-\frac{1}{4}\frac{1}{1-\frac{(x+1)^2}{4}}
$$ and then use the expansion 
$$
\frac{1}{1-u}=1+u+u^2+u^3+ \ldots 
$$ as $|u|<1.$ 
Hence
$$
\frac{1}{x^2+2x-3}=-\sum_{n=0}^{\infty}\frac{1}{4^{n+1}}(x+1)^{2n}, \quad |x+1|<2.
$$
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