General formula for Evaluating $\sum_{n=0}^\infty n^ar^n$ where $ |r|<1 , a\ge0$

Question

I'm trying to derive a general formula for $$\sum_{n=0}^\infty n^ar^n$$

where $a\ge0$ and $|r|<1$

I know the first couple a:

$$I(0)=\sum_{n=0}^\infty r^n=\frac{1}{(1-r)}$$

$$I(1)=\sum_{n=0}^\infty nr^n=\frac{r}{(1-r)^2}$$

$$I(2)=\sum_{n=0}^\infty n^2r^n=\frac{-r(r+1)}{(r-1)^3}$$

$$I(a)=\sum_{n=0}^\infty n^ar^n=???$$ assuming my math was correct. I got these by differentiating the general summation formula for geometric series. After doing it a couple more times, I can't seem to discern a pattern.

Answer 1

$$\sum_{n=0}^\infty n^a r^n=\frac{\sum_{m=0}^{a-1}A(a,m)x^{m+1}}{(1-r)^{a+1}}$$ for $a\ge1$, where the $A(a,m)$ are Eulerian numbers.

Answer 2

Mathematica yields $$\sum_{n=0}^{\infty}n^a \, r^n=\Phi(r,-a,0),$$ where $\Phi(z,s,a)$ is the Hurwitz-Lerch transcendant.