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

Question

I am trying to derive the generating function for $H(x,p) = \sum_{n=0}^{\infty} n^p x^n$

I am trying to solve it with the following logic:

(Edited now, trying a new framing)

Base case:

$$H(x,0) = \sum_{n=0}^{\infty} x^n = \dfrac{1}{1-x}$$

And

$$H(x,p) = x \dfrac{\partial}{\partial x} H(x,p-1)$$

Implying

$$H(x,p) = \left(x \dfrac{\partial}{\partial x}\right)^p \dfrac{1}{1-x}$$

Answer 1

Here's one way of describing the answer. First, the Stirling numbers of the second kind $S(m, k)$ have the property that

$$n^m = \sum_k S(m, k) (n)_k$$

where $(n)_k = n(n-1) \dots (n-(k-1))$. Second, the generating function of $(n)_k$ turns out to be relatively straightforward to write down: it is

$$\sum_n (n)_k x^n = k! \frac{x^k}{(1 - x)^{k+1}}.$$

You can derive this by repeatedly differentiating $\frac{1}{1 - x}$. It follows that

$$\sum_n n^m x^n = \sum_k S(m, k) k! \frac{x^k}{(1 - x)^{k+1}}.$$

This is in fact equivalent to a claim about how the differential operator $\left( x \frac{d}{dx} \right)^n$ expands, as mentioned in the comments.

Answer 2

Hint. An approach similar to what you are looking for. One may recall the general formula

$$ \sum_{n=0}^\infty f(n)x^n=\frac1{1-x}\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}\:\omega_n\left(\frac{x}{1-x} \right), \quad |x|<1, \tag1 $$

with the polynomials $$ \omega_n\left(x \right)=\sum_{k=0}^n \begin{Bmatrix}n\\k\end{Bmatrix}k!x^k \tag2 $$ where $\displaystyle \begin{Bmatrix}n\\k\end{Bmatrix}$ are the Stirling numbers of the second kind. The identity $(1)$ holds true for a vast class of regular functions $f$. A proof of $(1)$ can be found in Boyadzhiev's paper here.

Applying $(1)$ to $f(x)=x^p$, gives directly

$$ \sum_{n=0}^\infty n^px^n=\frac1{1-x}\:\omega_p\left(\frac{x}{1-x} \right), \quad |x|<1. \tag3 $$

Observe that we have $$ \left(x \frac{d}{dx} \right)^p\frac1{1-x}=\frac1{1-x}\:\omega_p\left(\frac{x}{1-x} \right).\tag4 $$