I am trying to find a series for $$f(z)=\frac{2}{(z+2)^2},$$ that's convergent for $|z-2|<4$.
I have found that \begin{align} f(z)&=\frac{2}{(z+2)^2} \\ &=-2 \ \frac{d}{dz}\left(\frac{1}{z+2}\right) \\ &=-2 \ \frac{d}{dz}\left(\frac{1}{4}\left(\frac{1}{1+\frac{z-2}{4}}\right)\right) \\ &=-2 \ \frac{d}{dz}\left(\sum_{n=0}^{\infty} (-1)^n\frac{(z-2)^n}{4^{n+1}}\right) \\ &=-2\sum_{n=1}^{\infty} (-1)^n\frac{n(z-2)^{n-1}}{4^{n+1}}. \end{align} I am unsure if this result is correct.