I am trying to figure out the solution to the power series

$$\sum_{k=0}^{\infty} k^n p^k$$

where $n$ is a positive integer and $0 \leq p < 1$.

I am sure the series is convergent. When plugging the first few values of $n$ into Wolfram Alpha it is able to come up with a solution very quickly, which indicates to me that there must be a standard way of solving this, however I have not been able to figure out how to proceed.

The first few solutions for specific $n$ are:

\begin{align} \sum_{k=0}^{\infty} k p^k &= \frac{p}{(p-1)^2} \\ \sum_{k=0}^{\infty} k^2 p^k &= \frac{p^2 + p}{-(p-1)^3} \\ \sum_{k=0}^{\infty} k^3 p^k &= \frac{p^3 + 4p^2 + p}{(p-1)^4} \\ \sum_{k=0}^{\infty} k^4 p^k &= \frac{p^4 + 11p^3 + 11p^2 + p}{-(p-1)^5} \\ \sum_{k=0}^{\infty} k^5 p^k &= \frac{p^5 + 26p^4 + 66p^3 + 26p^2 + p}{(p-1)^6} \end{align}


You may find a recurrence without using heavy machinery (a.k.a. special functions):

First, you write $S_n=\sum_{k=0}^\infty k^n p^k$, then differentiate $S_{n-1}$ w.r.t. $p$, you get:

$$\frac{\partial}{\partial p}S_{n-1}=\sum_{k=0}^\infty k^n p^{k-1}=\frac{1}{p}S_n$$

So $S_n=p\frac{\partial}{\partial p}S_{n-1}$. Furthermore, $S_0=\frac{1}{1-p}$


Let's start with

$$f(x)=\sum_{k=0}^\infty e^{kx}=\frac1{1-e^x}$$

Differentiate this $n$ times w.r.t. $x$, applying Faà di Bruno's formula,

$$f^{(n)}(x)=\sum_{k=0}^\infty k^ne^{kx}=\sum_{k=1}^n\frac{k!}{(1-e^x)^{1+k}}B_{n,k}(e^x,e^x,\dots,e^x)$$

where $B_{n,k}$ is an incomplete Bell polynomial with $n-k+1$ variables. Setting $e^x=p$, we get

$$\sum_{k=0}^\infty k^np^k=\sum_{k=1}^n\frac{k!}{(1-p)^{1+k}}B_{n,k}(p,p,\dots,p)$$

  • $\begingroup$ Like it a lot +1 $\endgroup$ – Isham Sep 11 '17 at 0:07
  • 1
    $\begingroup$ You know, I've always wondered if anything more can be said about $B_{n,k}(x,x,\dots,x)$ $\endgroup$ – Simply Beautiful Art Sep 11 '17 at 0:16

The coefficients are given by Euler's triangle.


That is, $$\sum_{k=0}^\infty k^n p^k = \frac{1}{(1-p)^{n+1}} \sum_{m=1}^n A(n,m-1) p^m$$ where $A(n,m)$ is the Eulerian number.


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