How to show $\gamma =\sum_{m=2}^{\infty}(-1)^{m}\frac{\zeta (m)}{m}$?

Question

How to show $$\gamma =\sum_{m=2}^{\infty}(-1)^{m}\frac{\zeta (m)}{m}$$ where $\gamma $ is the Euler-Mascheroni constant and $\zeta (m)$ is the Riemann Zeta Function.

Answer 1

Hint: Use the series definition of the zeta function, $\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$, to rewrite your sum as a double sum, and then change the order of summation.

$$\sum_{m=2}^{\infty}(-1)^{m}\frac{\zeta (m)}{m}=\sum_{m=2}^{\infty}\frac{(-1)^{m}}{m}\sum_{n=1}^{\infty}\frac{1}{n^m}\\ =\sum_{m=2}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{m}}{m\,n^m}\\ =\sum_{n=1}^{\infty}\sum_{m=2}^{\infty}\frac{(-1)^{m}}{m\,n^m}\\ =\sum_{n=1}^{\infty}\left(\frac{1}{n}-\log{\frac{n+1}{n}}\right)\\ =\gamma$$