General formula for the power sum $\sum_{k=1}^{n}k^mx^k\quad \forall\; m\in \mathbb{N}$

Question

In my last question, it turns out to be solving the formula of $\sum_{k=1}^{n}k\omega^k$. I am curious if there is a geranal formula for the power sum: $$\sum_{k=1}^{n}k^mx^k\quad \forall\; m\in \mathbb{N}$$ I have calculated first few formula. \begin{align} \sum_{k=1}^{n}x^k&=\frac{x}{x-1}(x^n-1)\\ \sum_{k=1}^{n}kx^k&=\frac{x}{x-1}\left(x^n\left(n-\frac{1}{x-1}\right)+\frac{1}{x-1}\right)\\ \sum_{k=1}^{n}k^2x^k&=\frac{x}{x-1}\left(x^n\left(n^2-\frac{2n-1}{x-1}+\frac{2}{(x-1)^2}\right)-\left(\frac{1}{x-1}+\frac{2}{(x-1)^2}\right)\right)\\ \sum_{k=1}^{n}k^3x^k&=\frac{x}{x-1}\left(x^n\left(n^3-\frac{3n^2-3n+1}{x-1}+\frac{6n-6}{(x-1)^2}-\frac{6}{(x-1)^3}\right)+\left(\frac{1}{x-1}+\frac{6}{(x-1)^2}+\frac{6}{(x-1)^3}\right)\right) \end{align} The formula has a common form. $$ \sum_{k=1}^{n}k^mx^k=\frac{x}{x-1}\left(x^nf_m\left(\frac{1}{x-1}\right)-\left.f_m\left(\frac{1}{x-1}\right)\right|_{n=0}\right)\\ f_m \text{ is a polynomial of } \frac{1}{x-1} $$

It seems there are some rules for $f_m\left(\frac{1}{x-1}\right)$when $m$ increased. For example:

The constant term coefficient of $f_m\left(\frac{1}{x-1}\right)$ is $\;n^m$.

The $\frac{1}{x-1}$ term coefficient of $f_m\left(\frac{1}{x-1}\right)$ is $\;(n-1)^m-n^m$.

The $\frac{1}{(x-1)^m}$ term coefficient of $f_m\left(\frac{1}{x-1}\right)$ is $\;(-1)^mm!$.

I also find a recursive formula for the power sum. $$\sum_{k=1}^{n}k^{m+2}x^k=\sum_{k=1}^{n}k^{m+1}x^k+n(n+1)\sum_{k=1}^{n}k^mx^k-\sum_{k=1}^{n}\left(2k\sum_{j=1}^{k}j^mx^j\right)$$ Then the formula of $(m+2)^{th}$ order can always be calculated from previous two formula of $(m+1)^{th}$ order and $m^{th}$ order.

I have calculated the formula up to $6^{th}$ order, but the calculations become extremely complex and tedious when the order increased.

The term $$\sum_{k=1}^{n}\left(2k\sum_{j=1}^{k}j^mx^j\right)$$ will be expanded to all the formula from order $1$ to order $m+1$, and each formula times a polynomial of $\frac{1}{x-1}$.

Is there a general formula like Faulhaber's formula to calculate the coefficients of the formula for this power sum in a easier way?

Answer 1

I do not understand why my post was deleted. Using the method in R. Mortini, Die Potenzreihe $\sum_{n} n^mx^n$, $m\in \mathbb N$ Elem. Math. 38 (1983), no.2, 49--51. Math. Rev. MR0700283 one immediately gets an explicit formula of the form $$\sum_{k=1}^n k^m x^k=\frac{1}{(1-x)^{m+1}}\left(\sum_{j=0}^m(a_j-b_{j,n})x^j +(1-x^{n+1})\;\sum_{j=0}^m b_{j,n} x^j \right),$$ where $a_0:=0$ and $$a_j=\sum_{i=0}^{j-1} (-1)^i {m+1\choose i}(j-i)^m$$ and $$b_{j,n}=\sum_{i=0}^{j} (-1)^i {m+1\choose i}(j-i+1+n)^m.$$ Just write $$\sum_{k=1}^n k^m x^k=\sum_{k=1}^\infty k^mx^k-x^{n+1}\sum_{k=0}^\infty (k+n+1)^m x^k.$$

Answer 2

Recalling the umbral calculus relation: $$ (x \frac{d}{dx})^m g(x)= \sum_{f=0}^{\infty} { m \brace f} \: x^f \: \frac{d^f}{dx^f} g(x) $$ where ${n \brace m} $ are the Stirling Numbers of the Second Kind.

One then has: $$ \sum_{k=1}^{n} k^m x^k = \sum_{f=0}^{\infty} { m \brace f} \: x^f \: \frac{d^f}{dx^f} \frac{1-x^{n+1}}{1-x} $$ Despite the sum going to infinity, the sum is finite for integer values $m$. For example the sum $$ \sum_{k=1}^{10} k^4 x^k = \left(\frac{1-x^{11}}{(1-x)^2}-\frac{11 x^{10}}{1-x}\right) x+7 \left(\frac{2 \left(1-x^{11}\right)}{(1-x)^3}-\frac{22 x^{10}}{(1-x)^2}-\frac{110 x^9}{1-x}\right) x^2+6 \left(\frac{6 \left(1-x^{11}\right)}{(1-x)^4}-\frac{66 x^{10}}{(1-x)^3}-\frac{330 x^9}{(1-x)^2}-\frac{990 x^8}{1-x}\right) x^3+\left(\frac{24 \left(1-x^{11}\right)}{(1-x)^5}-\frac{264 x^{10}}{(1-x)^4}-\frac{1320 x^9}{(1-x)^3}-\frac{3960 x^8}{(1-x)^2}-\frac{7920 x^7}{1-x}\right) x^4 $$ which, of course, is equivalent to $$10000 x^{10}+6561 x^9+4096 x^8+2401 x^7+1296 x^6+625 x^5+256 x^4+81 x^3+16 x^2+x $$