Polylog sums which evaluate to an integer

hbghlyj

For any $k\inZ$, $x\inZ$, show that $\displaystyle \sum_{n=1}^{\infty} n^{k-1} \frac{(x-1)^k}{x^n} \inZ$
E.g. $k = 15$, $x = 3$, the series evaluates to 696933753434112.

tommywong

Let $~p(n)=n^{k-1},~\deg(p(n))=k-1$

$\displaystyle f(n)
=\frac{p(n)}{1/x-1}+\frac{1}{(1/x-1)^2}\sum_{m=1}^{\deg(p)} \frac{ (-1)^m (1/x)^{m-1}}{(1/x-1)^{m-1}}\Delta^m(p(n))$
$\displaystyle =\frac{-xp(n)}{x-1}-\frac{x^2}{(x-1)^2}\sum_{m=1}^{k-1} \frac{ 1}{(x-1)^{m-1}}\Delta^m(p(n))$

$\displaystyle \sum_{n=1}^{\infty} n^{k-1} \frac{(x-1)^k}{x^n}
=-\frac{(x-1)^k}{x}f(0)$
$\displaystyle =(x-1)^{k-1}p(0)+x\sum_{m=1}^{k-1} (x-1)^{k-1-m}\Delta^m(p(0))$
$\displaystyle =x\sum_{m=1}^{k-1} (x-1)^{k-1-m}\Delta^m(p(0))$

hbghlyj

Sum[n^(k - 1) ((x - 1)^k/x^n),{n,1,oo}]$ = (x - 1)^k\text{Li}_{1 - k}\left(1\over x\right)$ when $\abs x>1$
$\text{Li}_n(x)$ is the polylogarithm function

tommywong

wolframalpha唔撚識差分算子，冇見佢用過