TITLE: A Tough Series $\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1}=-\gamma+\log(2)$
QUESTION [24 upvotes]: 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?

REPLY [4 votes]: This is similar in some respects, but different in others, from the answer provided by Shobhit Bhatnagar.  I also thought a detailed explanation of the steps might prove useful.
First we show a binocular telescoping series:
$$
\begin{align}
\sum_{n=2}^m\log\left(\frac{n^2}{n^2-1}\right)
&=\sum_{n=2}^m\log\left(\frac{n}{n-1}\right)+\sum_{n=2}^m\log\left(\frac{n}{n+1}\right)\tag{1a}\\
&=\log(m)+\log\left(\frac2{m+1}\right)\tag{1b}\\[3pt]
&=\log\left(\frac{2m}{m+1}\right)\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: $\frac{n^2}{n^2-1}=\frac{n}{n-1}\frac{n}{n+1}$
$\text{(1b)}$: evaluate $2$ telescoping sums
$\text{(1c)}$: simplify
Now we use $(1)$ to compute the sum:
$$
\begin{align}
&\sum_{k=1}^\infty\frac{\zeta(2k+1)-1}{k+1}\\
&=\sum_{k=1}^\infty\frac1{k+1}\sum_{n=2}^\infty\frac1{n^{2k+1}}\tag{2a}\\
&=\sum_{n=2}^\infty\sum_{k=1}^\infty\frac1{k+1}\frac1{n^{2k+1}}\tag{2b}\\
&=\sum_{n=2}^\infty\left(n\log\left(\frac{n^2}{n^2-1}\right)-\frac1n\right)\tag{2c}\\
&=\lim_{m\to\infty}\sum_{n=2}^m\left((m-(m-n))\log\left(\frac{n^2}{n^2-1}\right)-\frac1n\right)\tag{2d}\\
&=\lim_{m\to\infty}\left(m\log\left(\frac{2m}{m+1}\right)-\sum_{n=2}^m\sum_{j=n+1}^m\log\left(\frac{n^2}{n^2-1}\right)-(H_m-1)\right)\tag{2e}\\
&=\lim_{m\to\infty}\left(m\log\left(\frac{2m}{m+1}\right)-\sum_{j=3}^m\sum_{n=2}^{j-1}\log\left(\frac{n^2}{n^2-1}\right)-(H_m-1)\right)\tag{2f}\\
&=\lim_{m\to\infty}\left(m\log\left(\frac{2m}{m+1}\right)-\sum_{j=3}^m\log\left(\frac{2(j-1)}{j}\right)-(H_m-1)\right)\tag{2g}\\
&=\lim_{m\to\infty}\left(m\log\left(\frac{2m}{m+1}\right)-(m-2)\log(2)-\log\left(\frac2m\right)-(H_m-1)\right)\tag{2h}\\[6pt]
&=\lim_{m\to\infty}\left(m\log\left(\frac{m}{m+1}\right)+1+\log(2)+\log(m)-H_m\right)\tag{2i}\\[9pt]
&=\log(2)-\gamma\tag{2j}
\end{align}
$$
Explanation:
$\text{(2a)}$: expand $\zeta(2n+1)-1=\sum\limits_{k=2}^\infty\frac1{k^{2n+1}}$
$\text{(2b)}$: switch order of summation
$\text{(2c)}$: $n\log\left(\frac1{1-1/n^2}\right)-\frac1n=\sum\limits_{k=1}^\infty\frac1{(k+1)n^{2k+1}}$
$\text{(2d)}$: write infinite sum as a limit and $n$ as $m-(m-n)$
$\text{(2e)}$: apply $(1)$, the definition of the Harmonic Numbers, and $\sum\limits_{j=n+1}^m1=m-n$
$\text{(2f)}$: switch the order of summation
$\text{(2g)}$: apply $(1)$
$\text{(2h)}$: evaluate the telescoping sum
$\text{(2i)}$: cancel and combine terms
$\text{(2j)}$: evaluate the limits