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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?

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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 $f_{a-1}'(x)$.

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}}. $$

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

$$ \frac{k!}{ (1-x)^{k+1} }. $$

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