Have I found the correct Laurent series expansion? I am supposed to find the Laurent series expansion for
$$f(z) = \frac{z+1}{z^2-4}$$ in the region $1<|z+1|<3$. 
My solution:
$$w=z+1 \Leftrightarrow z=w-1 \Rightarrow f(z) = \frac{w-1+1}{(w-1)^2-4} = \frac{w}{(w+1)(w-3)}$$
Partial fractions: $$\frac{w}{(w+1)(w-3)} = \frac{1}{4}\frac{1}{w+1} + \frac{3}{4}\frac{1}{w-3}$$
We have: 
\begin{eqnarray}\frac{1}{4}\frac{1}{w+1} + \frac{3}{4}\frac{1}{w-3} &=& \frac{1}{4}\frac{1}{w} \bigg(\frac{1}{1+\frac{1}{w}}\bigg) + \frac{3}{4}\frac{1}{w}\bigg(\frac{1}{1-\frac{3}{w}}\bigg)\\ &=& \frac{1}{4}\frac{1}{w} \sum_{n=0}^\infty \bigg[\frac{(-1)}{w}\bigg]^n + \frac{3}{4}\frac{1}{w} \sum_{n=0}^\infty \bigg[\frac{3}{w}\bigg]^n\\ &=& \frac{1}{4} \sum_{n=0}^\infty \frac{(-1)^n}{w^{n+1}} + \frac{3}{4} \sum_{n=0}^\infty \frac{3^n}{w^{n+1}}\\ &=& \frac{1}{4} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{w^{n}} + \frac{3}{4} \sum_{n=1}^\infty \frac{3^{n-1}}{w^n}\\ &=& \frac{1}{4} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(z+1)^n} + \frac{3}{4} \sum_{n=1}^\infty \frac{3^{n-1}}{(z+1)^{n}}\end{eqnarray}
for the region $1<|z+1|<3$. It feels like there is something wrong here. Any help is greatly appreciated.

Very nice work! Unfortunately, this is valid on the region $|w|>3,$ rather than the one you want. In order to get $|w|<3,$ we must have $\left|\frac{w}{3}\right|<1,$ so instead, we'll rewrite $$\frac{1}{w-3}=-\frac13\frac1{1-\frac{w}{3}},$$ so that \begin{eqnarray}\frac14\frac1{w+1}+\frac34\frac1{w-3} &=& \frac{1}{4}\frac{1}{w} \sum_{n=0}^\infty \bigg[\frac{(-1)}{w}\bigg]^n - \frac{1}{4}\sum_{n=0}^\infty \bigg[\frac{w}{3}\bigg]^n\\ &=& \frac{1}{4} \sum_{n=0}^\infty \frac{(-1)^n}{w^{n+1}} - \frac{1}{4} \sum_{n=0}^\infty \frac{w^n}{3^n}\\ &=& \frac{1}{4} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{w^{n}} - \frac{1}{4} \sum_{n=1}^\infty \frac{w^n}{3^n}\\ &=& \sum_{n=1}^\infty \frac{(-1)^{n-1}}{4}w^{-n} - \sum_{n=1}^\infty \frac1{3^n\cdot 4}w^{n}\\ &=& \sum_{-\infty}^{n=-1} \frac{(-1)^{-n-1}}{4}w^n - \sum_{n=1}^\infty \frac1{3^n\cdot 4}w^{n}\\ &=& \sum_{-\infty}^{n=-1} \frac{(-1)^{n+1}}{4}w^n - \sum_{n=1}^\infty \frac1{3^n\cdot 4}w^{n}.\end{eqnarray}
Thus, we have $$\frac{z+1}{z^2-4}=\sum_{n\in\Bbb Z}a_n(z+1)^n,$$ where $$a_n=\begin{cases}-\cfrac1{3^n\cdot 4} & n\ge0\\-\cfrac{(-1)^n}{4} & n<0.\end{cases}$$
