Show that $\sum_{j=1}^{\infty}\dfrac{\zeta(6j-3)-1}{2j-1} =\frac12\ln(3/2) $ Show that
$\sum_{j=1}^{\infty}\dfrac{\zeta(6j-3)-1}{2j-1}
=\frac12\ln(3/2)
$.
This came out of
some work I did recently.
Any other context would be too much.
 A: Try switching the order of summation:
$$
\sum_{j=1}^{\infty}\frac{\zeta(6j-3)-1}{2j-1} = \sum_{j=1}^{\infty}\frac{\sum_{k=1}^{\infty}\frac{1}{k^{6j-3}}-1}{2j-1} 
$$
$$
= \sum_{j=1}^{\infty}\frac{1}{2j-1} \sum_{k=2}^{\infty}\frac{1}{k^{6j-3}}
$$
$$
=  \sum_{k=2}^{\infty}\sum_{j=1}^{\infty}\frac{1}{2j-1}\frac{1}{k^{6j-3}}
$$
The inner sum has a closed form:
$$
=  \sum_{k=2}^{\infty}\tanh(k^{-3})
$$
$$
= \frac{1}{2} \sum_{k=2}^{\infty}\log\left(\frac{1+k^{-3}}{1-k^{-3}}\right)
$$Turn this into an infinite product, which is easily seen to converge to $3/2$.
$$
= \frac{1}{2} \log\left(\prod_{k=2}^{\infty}\frac{1+k^{-3}}{1-k^{-3}}\right)
$$
$$
= \frac{1}{2} \log\left(3/2\right)
$$
A: Using integral representations
$$ \zeta(6j-3) = \frac{1}{(6j-4)!}\int_{0}^{+\infty}\frac{x^{6j-4}}{e^x-1}\,dx $$
$$ \zeta(6j-3)-1 = \frac{1}{(6j-4)!}\int_{0}^{+\infty}\frac{x^{6j-4}}{e^x(e^x-1)}\,dx $$
hence
$$\sum_{j\geq 1}\frac{\zeta(6j-3)-1}{2j-1}=\int_{0}^{+\infty}\frac{3}{xe^x(e^x-1)}\sum_{n\geq 0}\frac{x^{6n+3}}{(6n+3)!}\,dx $$
and by the DFT (Discrete Fourier Transform)
$$\sum_{n\geq 0}\frac{x^{3n}}{(3n)!}=\frac{1}{3}\left(e^x+e^{-x/2}\cos\frac{x\sqrt{3}}{2}\right) $$
$$\sum_{n\geq 0}\frac{x^{6n+3}}{(6n+3)!}=\frac{1}{3}\left(\sinh(x)-2\sinh\frac{x}{2}\cos\frac{x\sqrt{3}}{2}\right) $$
so in order to compute $\sum_{j\geq 1}\frac{\zeta(6j-3)-1}{2j-1}$ we may also invoke Frullani's integral, due to
$$ \sum_{j\geq 1}\frac{\zeta(6j-3)-1}{2j-1}= \int_{0}^{+\infty}\frac{1+e^x-2e^{x/2}\cos\frac{x\sqrt{3}}{2}}{2x e^{2x}}\,dx. $$
There is a massive simplification only due to the fact that two sixth roots of unity differ by one.
