How to derive this series for $\gamma$ that is only involving odd integer values of $\zeta(s)$? With $\gamma$ being the Euler Mascheroni constant, this series is well known:
$$1- \sum_{n=2}^{\infty} \frac{\zeta(n)-1}{n} = \gamma \tag{1}$$
The following series involving $\zeta(2n+1)$ also seems to converge to the same value, albeit slower:
$$1- \sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{(n+1)\,(2n+1)} = \gamma \tag{2}$$
Is there a way to derive (2) from (1) ?
 A: Note that
$$\frac{\zeta(2n+1)}{(n+1)\,(2n+1)}=2\frac{\zeta(2n+1)-1}{2n+1}-\frac{\zeta(2n+1)-1}{n+1}+\frac{2}{2n+1}-\frac{2}{2n+2}.$$
Hence
$$\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(n+1)\,(2n+1)}=
2\sum_{n=1}^{\infty}\frac{\zeta(2n+1)-1}{2n+1}-\sum_{n=1}^{\infty}\frac{\zeta(2n+1)-1}{n+1}+2\sum_{n=3}^{\infty}\frac{(-1)^{n-1}}{n}.$$
Finally we apply Question about series involving zeta function? and A Tough Series $\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1}=-\gamma+\log(2)$ :
$$\begin{align}\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(n+1)\,(2n+1)}&=
2\left(1-\gamma-\frac{\log(2)}{2}\right)-\left(-\gamma+\log(2)\right)+2\left(\log(2)-1+\frac{1}{2}\right)\\
&=\left(2-2\log(2)-\gamma\right)+2\log(2)-1\\
&= 1-\gamma.\end{align}$$
For a selfcontained proof, just note that
$$\begin{align}2\sum_{n=1}^{\infty}&\frac{\zeta(2n+1)-1}{2n+1}-\sum_{n=1}^{\infty}\frac{\zeta(2n+1)-1}{n+1}\\
&=\sum_{k=2}^\infty \sum_{n=1}^\infty \frac{2}{(2n+1)k^{2n+1}}-\sum_{k=2}^\infty\sum_{n=1}^\infty\frac{1}{(n+1)k^{2n+1}}\\
&=\sum_{k=2}^\infty \left(-\log \left(1-\frac{1}{k}\right)+\log \left( 1+\frac{1}{k}\right)-\frac{2}{k}\right)\\
&\qquad +\sum_{k=2}^\infty
\left(\frac{1}{k}+ k\log \left( 1+\frac{1}{k}\right)+ k\log \left( 1-\frac{1}{k}\right)\right)\\
&=\sum_{k=2}^\infty \left((k+1)\log \left( 1+\frac{1}{k}\right)
+(k-1)\log \left( 1-\frac{1}{k}\right)-\frac{1}{k}\right)\\
&=\sum_{k=2}^\infty \left((k+1)\log \left(k+1\right)-2k\log(k)
+(k-1)\log \left( k-1\right)-\frac{1}{k}\right)\\
&=\lim_{N\to \infty}\left((N+1)\log \left(N+1\right)
-2\log(2)-N\log(N)-H_N+1\right)\\
&=\lim_{N\to \infty}\left((N+1)\log \left(N+1\right)
-2\log(2)-N\log(N)-\log(N)-\gamma+1\right)\\
&=\lim_{N\to \infty}\left((N+1)\log \left(1+\frac{1}{N}\right)
-2\log(2)-\gamma+1\right)\\
&=2-2\log(2)-\gamma.
\end{align}$$
