How can I show that :$\sum_{n=1}^{\infty}(-1)^{n-1}(\zeta(n+1)-1))=\frac12$?

Question

A few computations I did with Mathematica gave me this sum. I'm really very interested to know how I can evaluate $$\sum_{n=1}^{\infty}(-1)^{n-1}(\zeta(n+1)-1))=\frac12$$ I have used Direchlet eta function to show it but that didn't succeed since the value of zeta here is related only to an integer.

Answer 1

\begin{eqnarray*} \sum_{n=1}^{\infty} (-1)^{n-1}(\zeta(n+1)-1) &=& \sum_{n=1}^{\infty} (-1)^{n-1} \sum_{m=2}^{\infty} \frac{1}{m^{n+1}} \\ \end{eqnarray*}\begin{eqnarray*}&=& \sum_{m=2}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{m^{n+1}} \\ \end{eqnarray*} \begin{eqnarray*}&=& \sum_{m=2}^{\infty} \frac{1}{m(m+1)}=\frac{1}{2} \\ \end{eqnarray*} Use partial fractions and telescoping to do the final sum.