Prove $\sum_{n=1}^\infty \frac{\zeta(2n+1) - 1}{(2n+1) 2^{2n}} = 1 + \ln(2) - \ln(3) - \gamma$ I've found the following series on the Wikipediapage of the Euler-Mascheroni constant and I want to prove it.
$$\sum_{n=1}^\infty \frac{\zeta(2n+1) - 1}{(2n+1) 2^{2n}} = 1 + \ln(2) - \ln(3) - \gamma$$
I know that $$\sum_{k=1}^\infty \frac{\zeta(2k) - 1}{k} = \ln(2)$$
Any help would be appreciated. Thanks in advance.
 A: First, we use the familiar series representation of the Riemann Zeta function to write
$$\zeta(2n+1)-1=\sum_{k=2}^\infty \frac{1}{k^{2n+1}}\tag1$$
Next, define the function $f(x)$ as
$$f(x)=\sum_{n=1}^\infty \frac{\zeta(2n+1)-1}{2n+1}\,x^{2n+1}\tag2$$
Using $(1)$ in $(2)$ reveals
$$\begin{align}
f(x)&=\int_0^x \sum_{n=1}^\infty \sum_{k=2}^\infty \frac{t^{2n}}{k^{2n+1}}\,dt\\\\
&=\int_0^x \sum_{k=2}^\infty \frac1k \sum_{n=1}^\infty \left(\frac{t^2}{k^2}\right)^n \,dt \\\\
&=\int_0^x \sum_{k=2}^\infty \frac1k \frac{t^2}{k^2-t^2}\,dt\\\\
&=\int_0^x \left(\frac{t^2}{t^2-1}+\sum_{k=1}^\infty \frac1k\frac{t^2}{k^2-t^2}\right)\,dt\\\\
&=\int_0^x \frac{t^2}{t^2-1}\,dt-\frac12\int_0^x \sum_{k=1}^\infty\left(\frac1k-\frac1{k+t}\right)\,dt-\frac12\int_0^x \sum_{k=1}^\infty\left(\frac1k-\frac1{k-t}\right)\,dt\\\\
&=\int_0^x \frac{t^2}{t^2-1}\,dt-\frac12 \int_0^x \left(2\gamma +\psi(t+1)+\psi(1-t)\right)\,dt\\\\
&=x+\log\left(\frac{1-x}{1+x}\right)-\gamma x -\frac12 \log\left(\frac{\Gamma(1+x)}{\Gamma(1-x)}\right)
\end{align}$$
Setting $x=1/2$ and multiplying by $2$ reveals
$$\sum_{n=1}^\infty \frac{\zeta(2n+1)-1}{(2n+1)2^{2n}}=1-\log(3)-\gamma+\log(2)$$
as was to be shown!
