How do we show that $\ln{2}-\gamma=\sum_{n=1}^{\infty}{\zeta(2n+1)\over 2^{2n}(2n+1)}?$ 
$$\ln{2}-\gamma=\sum_{n=1}^{\infty}{\zeta(2n+1)\over 2^{2n}(2n+1)}\tag1$$

Any hints?
 A: Hint. Due to normal convergence, one is allowed to write
$$
\begin{align}
\sum_{n=1}^{\infty}{\zeta(2n+1)\over 2^{2n}(2n+1)}&=\sum_{n=1}^{\infty}\left({1\over 2^{2n}(2n+1)}\sum_{k=1}^{\infty}\frac1{k^{2n+1}}\right)
\\\\&=\sum_{k=1}^{\infty}\left(\sum_{n=1}^{\infty}\frac{1}{2^{2n}(2n+1)}\frac1{k^{2n+1}}\right)
\\\\&=\sum_{k=1}^{\infty}\left(\!\ln\frac{2k+1}{2k-1}-\frac1k\!\right)
\\\\&=\lim_{N \to \infty}\sum_{k=1}^N\left(\ln(2k+1)-\ln(2k-1)-\frac1k\right)
\\\\&=\lim_{N \to \infty}\left(\!\ln(2N+1)-H_{N} \frac{}{}\!\right)
\\\\&=\lim_{N \to \infty}\left(\!\ln2 -(H_N-\ln N)+\ln\left(1+\frac1{2N}\right)\!\right)
\\\\&=\ln 2-\gamma.
\end{align}
$$ 
A: It is enough to exploit the integral representation
$$ \zeta(2n+1) = \frac{1}{(2n)!}\int_{0}^{+\infty}\frac{x^{2n}}{e^{x}-1}\,dx \tag{1} $$
to get:
$$ \sum_{n\geq 1}\frac{\zeta(2n+1)}{4^n(2n+1)} = \int_{0}^{+\infty}\frac{dx}{e^x-1}\sum_{n\geq 1}\frac{x^{2n}}{4^n(2n+1)!} = \int_{0}^{+\infty}\frac{-x+2\sinh\frac{x}{2}}{x(e^x-1)}\,dx\tag{2}$$
The last integral can be easily converted into two elementary integrals, giving $\gamma$ and $\log 2$.
For instance:
$$ \gamma=\sum_{n\geq 1}\left(\frac{1}{n}-\log\frac{n+1}{n}\right)=\int_{0}^{+\infty}\sum_{n\geq 1}\left(e^{-nx}-\frac{e^{-nx}-e^{-(n+1)x}}{x}\right)=\int_{0}^{+\infty}\left(\frac{1}{e^x-1}-\frac{e^{-x}}{x}\right)\,dx.\tag{3}$$
A: We can also prove the identity from $$\sum_{k\geq2}\frac{\left(-x\right)^{k}\zeta\left(k\right)}{k}=x\gamma+\log\left(\Gamma\left(x+1\right)\right),\,-1<x\leq1
 $$ which follows from the Weierstrass product of the Gamma function. So we can see that $$\sum_{k\geq1}\frac{\zeta\left(2k+1\right)}{2^{2k}\left(2k+1\right)}=2\sum_{k\geq1}\frac{\zeta\left(2k+1\right)}{2^{2k+1}\left(2k+1\right)}=2\left(\sum_{k\geq2}\frac{\zeta\left(k\right)}{2^{k}k}-\sum_{k\geq1}\frac{\zeta\left(2k\right)}{2^{2k}2k}\right)
 $$ and from $$\sum_{k\geq1}\zeta\left(2n\right)x^{2n}=\frac{1-\pi x\cot\left(\pi x\right)}{2}
 $$ we have $$\sum_{k\geq1}\frac{\zeta\left(2n\right)}{2^{2n}2n}=\frac{1}{2}\int_{0}^{1/2}\left(\frac{1}{x}-\cot\left(\pi x\right)\right)dx=\frac{\log\left(\frac{\pi}{2}\right)}{2}
 $$ so finally $$\sum_{k\geq1}\frac{\zeta\left(2k+1\right)}{2^{2k}\left(2k+1\right)}=-\gamma+2\log\left(\Gamma\left(\frac{1}{2}\right)\right)-\log\left(\frac{\pi}{2}\right)=\color{red}{-\gamma+\log\left(2\right)}
 $$ as wanted.
A: This is an immediate consequence of the Euler-Stieltjes series for $\gamma$: $$\gamma = 1 - \log\left(\frac{3}{2}\right) - \sum_{n = 1}^{\infty}\frac{\zeta(2n + 1) - 1}{2^{2n}(2n + 1)}\tag{1}$$ We just need to note that $$\sum_{n = 1}^{\infty}\frac{1}{2^{2n}(2n + 1)} = \sum_{n = 1}^{\infty}\frac{x^{2n}}{2n + 1} = \frac{1}{2x}\log\left(\frac{1 + x}{1 - x}\right) - 1\tag{2}$$ where $x = 1/2$ and hence the sum in $(2)$ is equal to $$\log 3 - 1$$ and then from $(1)$ we get $$\gamma = 1 - \log (3/2) - \sum_{n = 1}^{\infty}\frac{\zeta(2n + 1)}{2^{2n}(2n + 1)} + \log 3 - 1$$ or $$\sum_{n = 1}^{\infty}\frac{\zeta(2n + 1)}{2^{2n}(2n + 1)} = \log 2 - \gamma$$ A proof of Euler-Stieltjes series and its generalization is given in an excellent answer by user robjohn.
