$$S = \sum_{n=0}^{\infty} \frac{1}{3^n} \sum_{k=0}^{n} \frac{k}{2^k}$$
We know that if one signal y can be written as $$y(n) = \sum_{k=0}^{n} x(k)$$ then it's Z transform is $$ \frac{z}{z-1} X(z) $$ where $X(z)$ is the Z transform of signal x.
If we let $$x(n) = \sum_{k=0}^{n} \frac{k}{2^k}$$ we can calculate it's Z transform but I'm stuck on what to do next, the answer should be $?$ (my professor put wrong answer apparently)
I get a few different results. The $z$-transform $Y(z)$ of $y$ is $$\frac1{1-z}X(z).$$ The $z$-transform $\mathrm{Id}(z)$ of $\mathrm{id}:n \to n$ is $$\frac{z}{(1-z)^2}.$$ So the $z$-transform of the given sequence is $$\frac1{1-\frac{z}3}\frac{\frac{z}6}{(1-\frac{z}6)^2}.$$ Plug in $z=1$ to find the sum $\frac9{25}$.