I have done series with $\zeta(2k)$ and $\zeta(k)$, but I have no idea with this one:

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

$\gamma$ is the Euler–Mascheroni Constant.

This value was given by Mathematica. Any hint?

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    $\begingroup$ Expand $\zeta(2 k + 1)$ and interchange summation order? $\endgroup$ – vonbrand Apr 13 '13 at 14:06
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    $\begingroup$ @vonbrand I think you get a logarithmic sum by doing that. How do you evaluate that? $\endgroup$ – Ishan Banerjee Apr 13 '13 at 14:47
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    $\begingroup$ The same constant arises in $$lim_{n \to \infty} \left(log(2n+1)-H_n\right) = log(2)-\gamma$$ math.stackexchange.com/a/439184/134791 $\endgroup$ – Jaume Oliver Lafont Jan 10 '16 at 23:20

I solved it myself.

First we note that

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

Then $$\begin{aligned} \sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1} &=\sum_{n=2}^\infty \left( -\frac{1}{n}- n\log \left( 1-\frac{1}{n^2}\right)\right) \\ &= \lim_{N\to \infty}\sum_{n=2}^N \left( -\frac{1}{n}- n\log \left( 1-\frac{1}{n^2}\right)\right)\\ &= \lim_{N\to \infty} \left[ -H_N+1-\sum_{n=2}^N n \log(n^2-1)+2\sum_{n=2}^Nn\log(n)\right]\\ &= \lim_{N\to \infty} \left[ -H_N+1-\sum_{n=2}^N \left(n\log(n+1) +n\log(n-1)-2n\log(n)\right)\right] \\ &= \lim_{N\to \infty} \Bigg[ -H_N+1+\log(2)-\sum_{n=3}^{N+1}(n-1)\log(n)-\sum_{n=3}^{N-1}(n+1)\log(n) \\ &\quad+\sum_{n=3}^N2n\log(n)\Bigg] \\ &= \lim_{N\to \infty}\left[-H_N-N\log(N+1)-(N-1)\log(N)+2N\log(N)+1+\log(2) \right]\\ &= \lim_{N\to \infty}\left(- \left(H_N-\log N \right)+\log(2)+1-N\log \left( 1+\frac{1}{N}\right)\right)\\&= \lim_{N\to \infty}\left( - \left(H_N-\log N \right)+\log(2)+\mathcal{O}(N^{-1})\right) \end{aligned}$$

Since $\displaystyle \gamma=\lim_{N\to \infty}(H_N-\log(N))$, we get

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


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