For my 900 Followers and Comrades

For any integer $k \ge 0$ we have $\sum_{n=0}^\infty (-1)^{n+1}n^k x^n \; = \; \left(x\frac{d}{dx}\right)^k \left(\sum_{n=0}^\infty (-1)^{n+1}x^n\right) \; = \; -\left(x\frac{d}{dx}\right)^k \frac{1}{1+x} \qquad |x| < 1$ Putting $x = e^y$ for $y < 0$ we see that $\sum_{n=0}^\infty (-1)^{n+1}n^k x^n \; = \; -\left(\frac{d}{dy}\right)^k \frac{1}{1+e^y}$ and hence $A_k$ , the Abel sum of $\sum_{n=1}^\infty (-1)^{n+1}n^k$ is equal to $A_k \; = \; \lim_{x\to1-}\sum_{n=0}^\infty (-1)^{n+1}n^k x^n \; = \; -\lim_{y\to0-}\left(\frac{d}{dy}\right)^k \frac{1}{1+e^y}$ and hence $A_k$ is equal to $k!$ times the coefficient of $y^k$ in the Maclaurin expansion of $-\frac{1}{1+e^y}$ . Since $\frac{1}{1+e^y} \; =\; \tfrac12 - \sum_{m=0}^\infty \frac{(-1)^m(2^{2m+2}-1)}{2^{2m+1}\pi^{2m+2}}\zeta(2m+2)y^{2m+1}$ we deduce that $A_k = \boxed{0}$ for all even $k$ except $k=0$ , and we have an expression for $A_k$ for odd integers $k$ that can be written in terms of Bernoulli numbers.

$A_n=-Li_{-n}(-1)=-\eta(-n)=(1-2^{1+n})\zeta(-n)=\frac{(2^{1+n}-1)B_{n+1}}{n+1}$

where $Li$ is the polylogarithm, $\eta(s)$ is Dirichlet's eta function, $\zeta(s)$ is the zeta function, and $B_n$ are the Bernoulli numbers. Thus $A_n=\boxed{0}$ for positive even integers $n$ since the zeta function has its trivial zeroes at $-n$ .

See Comrade @Pi Han Goh 's inspirational problem for the first step.