After seing this question I started wondering about a generalization of a similar sum. The sum is $$ S(m,n)=\sum_{r=1}^{m}(-1)^{r-1}\;r^{n} $$
I gave this to WA to crunch and it gave $$ S(m,n)= (-1)^{m}2^{n}\left(\zeta\left(-n, \frac{m+2}{2}\right)-\zeta\left(-n,\frac{m+1}{2}\right) \right)-(2^{n}-1)\zeta(-n) $$
I know that Bernoully numbers,$B_{n}$, and Bernoully polynomials, $B_{n+1}(x)$, are involved expressing $$ \zeta(-n,x)=-\frac{B_{n+1}(x)}{n+1}\;\;\;and\;\;\;\zeta(-n)=-\frac{B_{n+1}}{n+1} $$ But I wonder if this expression could, somehow, be simplified to express $S(m,n)$ in a more pleasant way, involving somehow powers of two.
I think that when $m$ is a Mersenne number, i.e. of the form $2^{m}-1$, it will be possible to obtain some kind of oversimplification.
Anyways, what I'm looking for is some other forms to express $S(m,n)$ with a special interest in the case of the upper limit of the sum being $2^{m}-1$, i.e. in the case $$ \begin{align*} S(m,n)&=\sum_{r=1}^{2^{m}-1}(-1)^{r-1}\;r^{n}\\ &=-2^{n}\left(\zeta\left(-n, \frac{2^{m}+1/2}{2}+\frac{1}{4}\right)-\zeta\left(-n,\frac{2^{m}+1/2}{2}-\frac{1}{4}\right) \right)-(2^{n}-1)\zeta(-n) \end{align*}\\ $$
Thanks is advance.
Suppose $2|m$
Consider our sum $$S = \sum_{j = 1}^{m}{(-1)^{j + 1}j^{n}} = S(n,m) - 2\sum_{j = 1}^{m/2}{(2\cdot j)^{n}} = S(n,m) - 2\cdot 2^{n}S(n,\frac{m}{2})$$
If $m \equiv1 {\mod{2}} $, then :
$$S = S(n,m) - 2\cdot 2^{n}S(n,\frac{m - 1}{2})$$, actually I don't know what is the better simplification.
UPD: $S(n,m) = \frac{1}{n+1}\sum_{j = 0}^{n} (-1)^{j} \binom{n+1}{j}B_{j}m^{n+1 -j} $