Generating function for $\sum_{n=0}^{\infty} n^k x^n$

Question

I would like to get the closed form for this generating function, assuming the $k$ is given up front / held constant: $\sum_{n=0}^{\infty} n^k x^n$

However I don't know if this is too advanced for me or if I am in over my head because I can't seem to get anywhere with this. One thing I did find, playing around with derivatives, is that if:

$F_0(x) = \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$

$F_1(x) = xF'_0(x) = \sum_{n=0}^{\infty} n x^{n}$

$F_2(x) = xF'_1(x) = \sum_{n=0}^{\infty} n^2 x^{n}$

$F_3(x) = xF'_2(x) = \sum_{n=0}^{\infty} n^3 x^{n}$

$F_k(x) = xF'_{k-1}(x) = \sum_{n=0}^{\infty} n^k x^{n}$ and this is what I want to compute the closed form for

Does a closed form exist for this or do I have to compute this in $O(k)$ time by going through each level, computing the derivative, multiplying by $x$, and so on?

Answer 1

A closed form does exist. One has $$ \sum_{n=0}^\infty n^px^n=\frac1{1-x}\:\omega_p\left(\frac{x}{1-x} \right), \quad |x|<1, $$ with the polynomials $$ \omega_n\left(x \right)=\sum_{k=0}^n \begin{Bmatrix}n\\k\end{Bmatrix}k!x^k $$ where $\displaystyle \begin{Bmatrix}n\\k\end{Bmatrix}$ are the Stirling numbers of the second kind.

A proof can be found in Boyadzhiev's paper. A related formula: eulerian polynomials.