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?
 A: 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.
A: 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 $f_{a-1}'(x)$.
A: A related technique. Here is how you advance,
$$ \sum_{n=0}^{\infty}x^n = \frac{1}{1-x} \implies (xD)^a \sum_{n=0}^{\infty}x^n= (xD)^a \frac{1}{1-x}  $$
$$ \implies \sum_{n=0}^{\infty}n^a x^n = \sum_{k=0}^{a} {a \brace k} x^k D^k \frac{1}{1-x},$$
where $ {n \brace k} $ are the stirling numbers of the second kind. Now, you just need to find the $k$-th derivative of $\frac{1}{1-x}$ and then subs $x=\frac{1}{b}$.
Added: The final answer is 

$$  \sum_{n=0}^{\infty}n^a x^n = \sum_{k=0}^{a} {a \brace k} \frac{k!\left(\frac{1}{b}\right)^k}{ \left(1-\left(\frac{1}{b}\right)\right)^{k+1} },\quad b\neq 1. $$

Note: we used the identity

1)$$ (xD)^a=\sum_{k=0}^{a} {a \brace k} x^k D^k.$$
2) The $k$-th derivative of $\frac{1}{1-x}$ is 
  $$ \frac{k!}{ (1-x)^{k+1} }. $$

A: Let us consider the following series instead:
$$ S_{a}(x) = \sum_{n=0}^{\infty} n^{a} x^{n}. $$
If we define a linear operator $L_{k}$ by
$$ L_{k} = x^{k} \frac{d^{k}}{dx^{k}}, $$
it is easy to find that $ L_{1}L_{n} = n L_{n} + L_{n+1} $ holds. Thus by the mathematical induction together with the recursive formula for the Stirling's number of the second kind, it follows that
$$ L_{1}^{n} = \sum_{k=1}^{n} \left\{ {n \atop k} \right\} L_{k}. $$
Then for a positive integer $a$, we have
$$ S_{a}(x)
= L_{1}^{a} S_{0}(x)
= \sum_{k=1}^{a} \left\{ {a \atop k} \right\} L_{k} \frac{1}{1-x}
= \sum_{k=1}^{a} \left\{ {a \atop k} \right\} \frac{k! x^{k}}{(1-x)^{k+1}}. $$
