I was given this series:
Let $q \in \mathbb{C}, \mid q\mid <1. $
$$\frac{1}{2}\sum_{n=0}^{\infty} (n^2 +3n +2)q^n $$
Now I have to show that
$$\frac{1}{(1-q)^3}=\frac{1}{2}\sum_{n=0}^{\infty} (n^2 +3n +2)q^n$$
and proof whether or not this series converges absolutely.
For the convergence I guess I have to use one of the many convergence criteria and for the equality I have no clue how to even start.
Since$$\frac1{1-q}=1+q+q^2+q^3+\cdots,$$you have, by the Cauchy product,$$\begin{align}\frac1{(1-q)^2}&=\frac1{1-q}\times\frac1{1-q}\\&=\left(1+q+q^2+q^3+\cdots\right)\times\left(1+q+q^2+q^3+\cdots\right)\\&=1+2q+3q^2+4q^3+\cdots\\&=\sum_{n=0}^\infty(n+1)q^n.\end{align}$$Now, in order to prove what you want to prove, apply the same idea to the equality$$\frac1{(1-q)^3}=\frac1{1-q}\times\frac1{(1-q)^2}$$and use the fact that$$(n+1)+n+(n-1)+\cdots+1=\frac{(n+1)(n+2)}2=\frac{n^2+3n+2}2.$$