Convergence of Taylor series expansion

Question

The Taylor series expansion of $\frac{1}{(x^2 -1)^2}$ at $x=-1$ is given by

$\frac{1}{(x^2 -1)^2} =\frac{1}{4(x+1)^{2}}+\frac{1}{4(x+1)}$+$\frac{3}{16}$+$\frac{(x+1)}{8}$ $\ldots$

Is this series convergent? If so, how do I prove it?

Answer 1

Hint.

To find the Laurent series at $z=-1$, note that $$ \frac{1}{(z^2-1)^2}= \frac14 \left(\frac{1}{(z+1)^2} +\frac{1}{(z+1)} -\frac{1}{z-1}+\frac{1}{(z-1)^2} \right) $$

So you need to work with the last two terms.

$$ \frac{1}{z-1}=\frac{1}{z+1-2}=-\frac12\frac{1}{1-\frac{z+1}{2}} $$

You can then find a series for $z$ with $|z+1|<2$. Taking the derivative term by term, you can find the series for $\frac{1}{(z-1)^2}$