How can I use $\sum_{j=0}^{\infty} z^j=\frac{1}{1-z}$ to find a closed form expression for $\sum_{j=0}^{\infty} jz^j$?

Question

How can I use $\sum_{j=0}^{\infty} z^j=\frac{1}{1-z}$ to find a closed form expression for $\sum_{j=0}^{\infty} jz^j$?

I have shown $\sum_{j=0}^{\infty} z^j$ by considering $S(z)=\sum_{j=0}^{\infty} z^j$ then subtracting $zS(z)$ and factoring. But I am not sure how to use this to find a closed form for $\sum_{j=0}^{\infty} jz^j$.

Answer 1

Without differentiating, observe \begin{align} (1-z)\sum^\infty_{j=0}jz^j=\sum^\infty_{j=1}jz^j-z\sum^\infty_{j=1}jz^j = \sum^\infty_{j=1}jz^j-\sum^\infty_{j=2}(j-1)z^j = \sum^\infty_{j=1}z^j = \frac{1}{1-z}-1 = \frac{z}{1-z}. \end{align}

Answer 2

By stars and bars, the number of $(a,b,c,\ldots)\in\mathbb{N}^k$ such that $a+b+c+\ldots=n$ is given by $\binom{n+k-1}{k-1}$.
In particular $$ \left(1+x+x^2+\ldots\right)^k = \frac{1}{(1-x)^k} = \sum_{n\geq 0}\binom{n+k-1}{k-1} x^n\tag{1}$$ and by considering the instances $k=1$ and $k=0$ we have $$ \sum_{n\geq 0}nx^n = \frac{1}{(1-x)^2}-\frac{1}{1-x} = \frac{x}{(1-x)^2}.\tag{2} $$