The following expressions for $\zeta(s)$ are valid for $s \in \mathbb{C}-\{1\}$:
By Helmut Hasse (1930):
$$\zeta(s)=\frac{1}{s-1}\sum_{n=0}^\infty\frac{1}{n+1}\sum_{k=0}^n(-1)^k\ {n \choose k}\ \frac{1}{(k+1)^{s-1}}$$
By Joseph Ser (1926):
$$\zeta(s)=\frac{1}{s-1}\sum_{n=0}^\infty\frac{1}{n+2}\sum_{k=0}^n(-1)^k\ {n \choose k}\ \frac{1}{(k+1)^{s}}$$
Their equivalence is nicely explained here.
Made a tweak to the latter series and numerical evidence suggests that:
$$f(s)=\sum_{n=1}^\infty\frac{1}{n}\sum_{k=0}^n(-1)^k\ {n \choose k}\ \frac{1}{(k+1)^{s}}$$
yields $f(s)=s$ for all $s$ (convergence is quite slow, but seems faster for negative $s$ and particularly at negative integers).
If true, why would this be the case?
ADDED:
Or formulated differently, is it true that:
$$f(s)=\lim_{N \to +\infty} H\left( N \right)+\sum _{n=1}^{N} \left( { \frac {1}{n}\sum _{k=1}^{n}{\left( -1 \right) ^{k}{n\choose k}\frac { 1 }{ \left( k+1 \right) ^{s}}}} \right) = s$$
with $H(N)$ = the $N$-th Harmonic Number?