Laurent series of $~\frac{1}{z+2}+\frac{1}{z^2}, ~~~~~~~0<|z+2|<2~$ Laurent series of $$~\frac{1}{z+2}+\frac{1}{z^2}, ~~~~~~~0<|z+2|<2~$$
I can't find a way to represent $z^2$ in terms of $z+2$. I've tried to do $(z+2)(z-2)+4$, but I'm stuck with the $+4$. 
 A: The Laurent series of $\frac1{z^2}$ about $-2$ is simply its Taylor series, which is$$\frac14+\frac28(z+2)+\frac3{16}(z+2)^2+\cdots+\frac{n+1}{4^{n+2}}(z+2)^n+\cdots$$So, the Laurent series that you're after is$$\frac1{z+2}+\frac14+\frac28(z+2)+\frac3{16}(z+2)^2+\cdots+\frac{n+1}{4^{n+2}}(z+2)^n+\cdots$$
A: Thanks for the answer ! 
I've tried the Taylor series, but because I was always simplifying the terms $\frac{2}{8}= \frac{1}{4}, \frac{4}{32}= \frac{1}{8}$ etc. to get it to look like the one in wolframalpha. I couldn't see the pattern to make the series looks like : $$\sum_{n=0}^\infty \frac{(n+1)}{2^{n+2}}(z+2)^n$$
Another solution is to do the Laurent series for $$\frac{1}{z} = \frac{1}{-2+(z+2)}$$ which is: $$-\sum_{n=0}^\infty \frac{1}{2^{n+1}}(z+2)^n$$ and take the derivative of both sides:
$(\frac{1}{z})^l = -\frac{1}{z^2}$ ; $(-\sum_{n=0}^\infty \frac{1}{2^{n+1}}(z+2)^n)^l = -\sum_{n=0}^\infty \frac{n+1}{2^{n+2}}(z+2)^n $ , multiplying by $-1$ and adding $\frac{1}{z+2}$, should look like:
$$\frac{1}{z+2}+\sum_{n=0}^\infty \frac{n+1}{2^{n+2}}(z+2)^n$$
