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?
 A: 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}
A: Hint: On $|z| < 1$, write $\frac{z^{2}}{1- z} = \sum_{n=0}^{\infty} z^{n +2}$. Differentiate twice and divide both sides by $z$ and plug in $z_{0} = \frac{i}{2}$ and you will get the right hand side to be the sum in question while the left hand side can be obtained via a careful working out of quotient rules of calculus. 
A: To make it more general consider $$S=\sum_{n=0}^\infty (n+a)\,(n+b)\, z^{n+c}$$ in which $a,b,c$ are just numbers (integer, rational, irrational or even complex). 
Start writing $$(n+a)\,(n+b)=An(n-1)+B(n-1)+C$$ Expanding and grouping terms, we have $$(a b+B-C)+n (a+b+A-B)+(1-A) n^2=0$$ So $$A=1\quad \, \quad B=1+a+b\quad \, \quad C=1 + a + b + a b$$ So, $$S=A\sum_{n=0}^\infty n(n-1)\, z^{n+c}+B\sum_{n=0}^\infty (n-1)\, z^{n+c}+C\sum_{n=0}^\infty \, z^{n+c}$$ $$S=Az^{c+2}\sum_{n=0}^\infty n(n-1)\, z^{n-2}+Bz^{c+1}\sum_{n=0}^\infty (n-1)\, z^{n-1}+Cz^c\sum_{n=0}^\infty \, z^{n}$$ where now we can recognize the sum of the geometric progression and its first and second derivatives.
The required expressions have been given in the answers.
