Simple way to evaluate $f(n)= \sum_{r=0}^{\infty}r^n x^r$

Question

Does anyone know a simpler formula than the one below for calculating values of this function for any positive integer n?

$$f(n)= \sum_{r=0}^{\infty}r^n x^r$$ Here's the derivation for the best formulae I could come up with: $$ \frac {1}{1-x} = \sum_{r=0}^{\infty} x^r$$ Differentiating n times: $$ \frac {n!}{(1-x)^{n+1}} = \sum_{r=0}^{\infty}(r+1)(r+2)...(r+n) x^r=\sum_{r=0}^{\infty} \frac 1r r^{(n+1)} x^r$$ Where $r^{(n)}$denots rising factorial. Expanding rising factorial in terms of Unsigned Stirling numbers of the first kind: $$ \frac {n!}{(1-x)^{n+1}} = \sum_{r=0}^{\infty}\frac 1r \sum_{k=1}^{n+1}c(n+1,k)r^kx^r=\sum_{r=0}^{\infty} \sum_{k=0}^{n}c(n+1,k+1)r^kx^r \\ \therefore \sum_{r=0}^{\infty}c(n+1,n+1)r^nx^r=\frac {n!}{(1-x)^{n+1}}-\sum_{r=0}^{\infty} \sum_{k=0}^{n-1}c(n+1,k+1)r^kx^r $$ Since $c(n+1,n+1)=1$, $$ f(n)=\frac {n!}{(1-x)^{n+1}}-\sum_{k=0}^{n-1}c(n+1,k+1)f(k)$$ This is a recurrence relation with $f(0)=\frac {1}{1-x}$ and thus should be solvable but I don't have the skill to do so.

Also, this function can be represented as a polylogarithm and this leads to a closed-form but still complicated solution in terms of Stirling numbers of the second kind: $$f(n)=Li_{-n}(x)=\sum_{r=0}^{\infty} (-1)^{n+r}S(n+1,r+1)\frac{r!}{(1-x)^{r+1}}$$

Any help solving the recurrence or exploiting properties of the polylogarithm would be very much appreciated.

ALSO solving the recurrence would lead to a relation between Stirling numbers of the first and second kinds and a different representation of the polylogarithm.

Answer 1

A slightly different approach is as follows: Let $$ S(t) = \sum_{r=0}^\infty e^{rt}x^r = \sum_{r=0}^\infty (e^t x)^r = \frac{1}{1-e^t x}, $$ for sufficiently small $x$. Then $$ S^{(n)}(t) = \frac{\partial^n}{\partial t^n} \left[ \frac{1}{1-e^t x} \right] = \frac{e^t x}{(1-e^t x)^{n+1}} \sum_{k=0}^{n-1} A(n,k+1) (e^t x)^k, $$ where $A(n,k+1)$ are the Eulerian numbers. Now, set $t=0$ to obtain $$ S^{(n)}(0) = f(n) = \sum_{r=0}^\infty r^n x^r = \frac{x}{(1-x)^{n+1}} \sum_{k=0}^{n-1} A(n,k+1) x^k = \frac{1}{(1-x)^{n+1}} \sum_{k=0}^{n-1} A(n,k+1) x^{n-k}. $$

I don't know if this should be considered 'simpler' than your formula, but at least it rids you of recurrences and infinite sums.

Answer 2

I'm not sure if this counts as an answer, and maybe you know this already, but here is what I think is a simpler recurrence relation:

$$f(n)= \begin{cases} \frac{1}{1-x},& \text{if }n=0, \\\\ \frac{x}{1-x}\sum_{k=0}^{n-1} \binom{n}{k} \,f(k),& \text{if } n\ge 1. \\\\ \end{cases}$$

I haven't thought through the convergence issues here carefully, but I think this is good for $\,\vert x \vert \lt 1.$

Proving the recurrence relation for $n \ge 1$ is a straightforward application of the binomial theorem.

\begin{align} \frac{x}{1-x}\sum_{k=0}^{n-1} \binom{n}{k} \,f(k) &= \frac{x}{1-x}\sum_{k=0}^{n-1} \binom{n}{k} \sum_{r=0}^\infty r^k x^r \\\\&=\frac{x}{1-x} \sum_{r=0}^\infty \sum_{k=0}^{n-1} \binom{n}{k} r^k \cdot x^r \\\\&= \frac{x}{1-x} \sum_{r=0}^\infty \left((r+1)^n-r^n \right)x^r \\\\&= \frac{x}{1-x} \left(\sum_{r=0}^\infty (r+1)^n x^r - \sum_{r=0}^\infty r^n x^r\right) \\\\&= \frac{x}{1-x} \left(\sum_{r=1}^\infty r^n x^{r-1} - \sum_{r=0}^\infty r^n x^r\right) \\\\&=\frac{x}{1-x} \sum_{r=0}^\infty r^n x^{r-1} (1-x) \\\\&=\sum_{r=0}^\infty r^n x^r \\\\&=f(n). \end{align}

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I don't know if it would be useful or not, but I'll throw in another formula here that might come in handy: $$\sum_{n=0}^\infty \frac{f(n)}{n!} = \frac{1}{1-ex},$$

proved as follows:

\begin{align} \sum_{n=0}^\infty \frac{f(n)}{n!} &=\sum_{n=0}^\infty \sum_{r=0}^\infty \frac{r^n x^r}{n!} \\\\&=\sum_{r=0}^\infty \sum_{n=0}^\infty \frac{r^n}{n!} x^r \\\\&=\sum_{r=0}^\infty e^r x^r \\\\&=\frac{1}{1-ex}. \end{align}

As before, I haven't really thought through the convergence issues, but I think this is valid for $\,\vert x \vert < 1/e.$