Calculate $\sum_{n=0}^\infty$ $(n+1)(n+2)(\frac{i}{2})^{n-1}$ I want to calculate $\sum_{n=0}^\infty$ $(n+1)(n+2)(\frac{i}{2})^{n-1}$.
I tried to separate it into a sum of real numbers ($n=0,2,4,\dots$) and complex numbers that are not real numbers ($n=1,3,5,\dots$) but it didn't work.
So I did it another way, using Cauchy's integral theorem:
Let $f(z)=(\frac{z}{2})^{n+2}$. Then $4f''(i)$= $(n+1)(n+2)(\frac{i}{2})^{n-1}$, which is a term of the sum I started with. I don't know how to proceed from here.
What can I do? How do I solve this?

Taking the derivative of the geometric series twice and dividing by $z$ gives
\begin{align}
\frac{1}{1-z} &= \sum_{n=0}^\infty z^n
\\
\frac{1}{(1-z)^2} &= \sum_{n=1}^\infty nz^{n-1}
\\
\frac{2}{(1-z)^3} &= \sum_{n=2}^\infty n(n-1)z^{n-2} = \sum_{n=0}^\infty (n+2)(n+1) z^n
\\
\frac{2}{z(1-z)^3} &= \sum_{n=0}^\infty (n+2)(n+1) z^{n-1}
\end{align}
