Proving this identity $\gamma=1+\ln(\frac{1}{2})+\sum_{k=1}^{\infty}(-1)^{k+1}\frac{\zeta(k+1)-1}{k+1}$ where $\gamma$ is the Euler-Masceroni constant

Question

I've seen this identity here

$$ \displaystyle \gamma=1+\ln(\frac{1}{2})+\sum_{k=1}^{\infty}(-1)^{k+1}\dfrac{\zeta(k+1)-1}{k+1} $$

and I'd like to know how it is deduced.

Could anyone help?

Thanks.

Answer 1

It's just a change of order of summation:

$$\begin{align} \sum_{k=1}^\infty (-1)^{k+1}\frac{\zeta(k+1)-1}{k+1} & = \sum_{k=1}^\infty \sum_{n=2}^\infty \frac{(-1)^{k+1}}{(k+1)n^{k+1}}\\ &= \sum_{n=2}^\infty\sum_{k=1}^\infty\frac{(-1)^{k+1}}{(k+1)n^{k+1}}\\ &= \sum_{n=2}^\infty \left(\frac1n - \log \left(1+\frac1n\right)\right) \end{align}$$

Now take a partial sum of the latter to obtain

$$H_N - \log (N+1) - 1 + \log 2 \xrightarrow[N\to\infty]{} \gamma - 1 + \log 2.$$

The change of order of summation is legitimate since the sum converges absolutely.