Summation of series $\sum_{n=1}^\infty \frac{n^a}{b^n}$?

How can we evaluate this series $$\sum_{n=1}^\infty \frac{n^a}{b^n}?$$

Here $a$ and $b$ are positive integers. If $b=1$ then series will be diverging, in other cases, it will be converging, but how to find this sum?

Isn't there a simple solution which doesn't involve Stirling numbers?

This may be rewritten as a polylogarithm : $$\sum_{n=1}^\infty \frac {n^a}{b^n}=\sum_{n=1}^\infty \frac {\left(\frac 1b\right)^n}{n^{-a}}=\operatorname{Li}_{-a}\left(\frac 1b\right)$$

For positive integers $a$ you may use these formulas that is compute successive derivatives of $\frac z{1-z}$. More exactly : $$\operatorname{Li}_{-a}(z)=\left(z\frac{\partial}{\partial z}\right)^a\frac z{1-z}=\sum_{k=0}^a k!\;S(a+1,k+1)\left(\frac z{1-z}\right)^{k+1}$$ with $S(n,k)$ the Stirling numbers of the second kind (replace $z$ by $\frac 1b$ in the final formula).

To see why this works begin with $a=0$ and note $z:=\frac 1b$ then we simply want the sum of a geometric series :

\begin{align} \operatorname{Li}_0(z)&=\sum_{n=1}^\infty z^n=\frac z{1-z}\\ \operatorname{Li}_{-1}(z)&=\sum_{n=1}^\infty n\,z^n=\left(z\frac{\partial}{\partial z}\right)\sum_{n=1}^\infty z^n=\frac z{(1-z)^2}\\ \operatorname{Li}_{-2}(z)&=\sum_{n=1}^\infty n^2\,z^n=\left(z\frac{\partial}{\partial z}\right)\sum_{n=1}^\infty n\,z^n=\frac {z(1+z)}{(1-z)^3}\\ \end{align} (and so on...)

Another nice formula is : $$\operatorname{Li}_{-a}(z)=\frac 1{(1-z)^{a+1}}\sum_{k=0}^{a-1} E(a,k)\, z^{a-k}$$ with $E(a,k)$ the Eulerian numbers.

hint

Write $x=1/b$. For $a=0$, this is a geometric series $\sum_{n=1}^\infty x^n$. Recursively, try to evaluate $$ f_a(x) = \sum_{n=1}^\infty n^a x^n $$ by noting that it is related to the derivative $f_{a-1}'(x)$.