Finding a value of a series using z-transform

Question

Having the z-transform formula: \begin{equation} V(x)=\sum_{n=0}^\infty x_nz^{-n} \end{equation}

I want to find the value of the series

\begin{equation} \sum_{n=0}^\infty \frac{n}{3^n} \end{equation}

we have that $z=3$ and $x_n=n$, using the table of transforms we have:

\begin{equation} \begin{array}{cc} n=\frac{z}{(z-1)^2} \\ 3^n=\frac{z}{z-3} \end{array} \end{equation}

thus the z-transform of the series is: \begin{equation} x(z)= \frac{z}{(z-1)^2}\frac{z}{z-3} \end{equation}

But how do I find the value of that series with this z-transform?

I tried using integration from $0$ to infinity, but that didn't give any convergent value.

How do I find that value of the series?

Thanks!

Answer 1

You do not need to transform $3^n$.

You have that $V(z) = \sum_{n=0}^\infty nz^{-n}$, and you are looking for $V(3)$.

The "table of transform" gives you that $V(z) = \frac{z}{(z-1)^2}$. Can you continue from here?