$$\sum _{ s=1 }^{ \infty }{ \left( \frac { 1 }{ 4s-1 } \sum _{ n=0 }^{ \infty }{ \left( \frac { 1 }{ n+1 } \sum _{ k=0 }^{ n }{ \left( \left( \begin{matrix} n \\ k \end{matrix} \right) \frac { { \left( -1 \right) }^{ k } }{ { \left( k+1 \right) }^{ s-1 } } \right) } \right) } -1 \right) }$$
I am getting it to be equal to $\displaystyle \sum_{s=1}^{\infty}{\left(\frac{(s-1)\zeta(s)}{4s-1} - 1\right)}$.
But the next sum seems to be diverging. Where am I wrong?
I reached till here by using $\displaystyle \int_{0}^{\infty}{{x}^{s-2} {e}^{-(k+1)x} \ dx} = \frac{(s-2)!}{{(k+1)}^{s-1}}$
We have: $$\begin{eqnarray*} \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{(k+1)^{s-1}} &=& \int_{0}^{+\infty}\frac{x^{s-2}}{(s-2)!}\sum_{k=0}^{n}\binom{n}{k}(-1)^k e^{-(k+1)x}\,dx\\&=&\int_{0}^{+\infty}\frac{x^{s-2}}{(s-2)!}e^{-x}(1-e^{-x})^n\,dx\end{eqnarray*}$$ hence: $$\begin{eqnarray*} \sum_{n\geq 0}\frac{1}{n+1}\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{(k+1)^{s-1}}&=&-\int_{0}^{+\infty}\frac{x^{s-1}}{(s-2)!(e^x-1)}\,dx\\&=&(1-s)\,\zeta(s)\end{eqnarray*}$$ and the original series is not converging.