# 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?

#### Solutions Collecting From Web of "Summation of series $\sum_{n=1}^\infty \frac{n^a}{b^n}$?"

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 $x=\frac{1}{b}$.

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