I have asked [this question](https://math.stackexchange.com/questions/3644735/why-does-this-tweak-to-sers-series-expression-for-zetas-always-yield-fs) at math stack exchange, however it did not get any traction. Still curious about the answer though.


Numerical evidence suggests that:


$$\lim\_{N \to +\infty} \sum\_{n=1}^N\frac{1}{n}\sum\_{k=0}^n(-1)^k\ {n \choose k}\ \frac{1}{(k+1)^{s}}=s$$


or equivalently


$$\lim\_{N \to +\infty} H\left( N \right)+\sum \_{n=1}^{N} \left( {
\frac {1}{n}\sum \_{k=1}^{n}{\left( -1 \right) ^{k}{n\choose k}\frac { 1
}{ \left( k+1 \right) ^{s}}}} \right) = s$$


with $H(N)$ = the $N$-th [Harmonic Number](https://en.wikipedia.org/wiki/Harmonic_number).


Convergence is quite slow, but clearly goes faster for negative $s$. Also, the computations for non-integer values of $s$ require high accuracy settings (I have used Maple, pari/gp and ARB).


However, according to Mathematica the series diverges by the "Harmonic series test", although when taking $s$ as an integer, it agrees on convergence. 


**Does this series converge for $s \in \mathbb{C}$ ?**


Some numerical results below:



```
s=0.5
0.497702121, N = 100
0.499804053, N = 1000
0.499905919, N = 2000

s=-3.1415926535897932385   
-3.14160222, N = 100
-3.14159284, N = 1000
-3.14159272, N = 2000

s=2.3-2.1i
2.45310498 - 1.94063637i, N = 100
2.33501943 - 2.09308517i, N = 1000
2.31996958 - 2.09923503i, N = 2000

```


Does this series, related to the Hasse/Ser series for $\zeta(s)$, converge for all $s \in \mathbb{C}$?
We have to show that 
$$l(s):=\sum\_{n=1}^\infty\frac1n\,S\_n(s)=s,$$
where 
$$S\_n(s):=\sum\_{k=0}^n(-1)^k\binom nk\frac1{(k+1)^s} \\
=\sum\_{k=0}^n(-1)^k\binom nk\int\_0^\infty du\,u^{s-1}e^{-(k+1)u}/\Gamma(s) \\
=\int\_0^\infty du\,u^{s-1}e^{-u}\sum\_{k=0}^n(-e^{-u})^k\binom nk/\Gamma(s) \\
=\int\_0^\infty du\,u^{s-1}e^{-u}(1-e^{-u})^n/\Gamma(s).$$
So, 
$$l(s)=\int\_0^\infty du\,u^{s-1}e^{-u}\sum\_{n=1}^\infty\frac1n\,(1-e^{-u})^n/\Gamma(s) \\
=\int\_0^\infty du\,u^s e^{-u}/\Gamma(s)
=\Gamma(s+1)/\Gamma(s)
=s,$$
as desired. 


