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?
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?
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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 $$ \frac{1}{x^2+2x-3}=-\sum_{n=0}^{\infty}\frac{1}{4^{n+1}}(x+1)^{2n}, \quad |x+1|<2. $$