$\sum\limits_{y=1}^{\infty}y\frac{\theta^ye^{-\theta}}{y!(1-e^{-\theta})}$

Question

Simplify the series: $$\sum_{y=1}^{\infty}y\frac{\theta^ye^{-\theta}}{y!(1-e^{-\theta})}$$

The solution given is $$\frac{\theta}{1-e^{-\theta}}\left(\sum_{z=0}^{\infty}\frac{\theta^ze^{-\theta}}{z!}\right)=\frac{\theta}{1-e^{-\theta}}$$

I do not understand how the solution is formed. Hope someone could explain it. Thanks in advance.

Answer 1

$$\begin{align} \sum_{y=1}^{\infty}y\frac{\theta^ye^{-\theta}}{y!(1-e^{-\theta})} &=\frac{e^{-\theta}}{(1-e^{-\theta})}\sum_{y=1}^{\infty}y\frac{\theta^y}{y!} \\[2ex] &=\frac{\theta e^{-\theta}}{(1-e^{-\theta})}\sum_{y=1}^{\infty}\frac{\theta^{y-1}}{(y-1)!} \\[2ex] &=\frac{\theta e^{-\theta}}{(1-e^{-\theta})}\sum_{y=0}^{\infty}\frac{\theta^{y}}{y!} \end{align}$$