Does $\sum\limits_{k \geq 0} \frac{1}{(4k+1)(4k+2)} = \frac{\log(2)}{4} + \frac{\pi}{8}$ hold? Context: I'm interested in trying to find as many explicit formulae as possible for:
$$ S_n(s) = \sum_{k \geq 0} \frac{1}{\prod_{i \in s} (nk+i)} $$
for all $ n \geq 2 $ and for all $ s $ subset of $ \{ 1, \dots, n \} $ such that $ \mbox{Card}(s) = 2 $.
So far, I can establish
$$ S_2(\{ 1, 2 \}) = \log(2) $$
$$ S_3(\{ 1, 2 \}) = \frac{\pi \sqrt{3}}{9},
S_3(\{ 2, 3 \}) = \frac{\log\left(3\right)}{2}-\frac{\pi\sqrt{3}}{18},
S_3(\{ 1, 3 \}) = \frac{\log\left(3\right)}{4}+\frac{\pi\sqrt{3}}{36} $$
$$ S_4(\{ 1, 3 \}) = \frac{\pi}{8}, S_4(\{ 2, 4 \}) = \frac{\log(2)}{4} $$
But I just guess that
$$ S_4(\{ 1, 2 \}) = \frac{\log(2)}{4} + \frac{\pi}{8} $$
Is there a generic method to establish all formulae?
 A: $$S=\sum_{0}^{\infty} \frac{1}{(4k+1)(4k+2)}=\sum_{k=0}^{\infty}\left(\frac{1}{4k+1}-\frac{1}{4k+2}\right)$$ $$=\sum_{k=0}^{\infty}\int_{0}^{\infty} [e^{-(4k+1)x}-e^{-(4k+2)x}]=\int_{0}^{\infty} \left( \frac{e^{-x}}{1-e^{=4x}}-\frac{e^{-2x}}{1-e^{-4x}}\right) dx $$ $$=\int_{0}^{\infty} \frac{e^{-x}(1-e^{-x})}{1-e^{-4x}}=\int_{0}^{1}\frac{(1-t)}{1-t^4}dt$$ $$=\frac{1}{2}\int_{0}^{1} \left(\frac{1}{1+t}+\frac{1}{1+t^2}-\frac{2t}{2(1+t^2)}\right) dt=\frac{1}{4}\ln 2+\frac{\pi}{8}.$$
A: Since $\displaystyle \frac{1}{4k+1} = \int_0^1 y^{4k}\, \mathrm{dy}$ and since $\displaystyle \frac{1}{4k+2} = \int_0^1 x^{4k+1}\, \mathrm{dx}$, we have:
\begin{aligned} S_4(\{ 1, 2 \}) & = \sum_{k \ge 0} \frac{1}{(4k+1)(4k+2)} \\& = \sum_{k \ge 0} \int_0^1\int_0^1y^{4k}x^{4k+1}\, \mathrm{dy} \, \mathrm{dx} \\& =  \int_0^1\int_0^1 \sum_{k \ge 0} y^{4k}x^{4k+1}\, \mathrm{dy} \, \mathrm{dx} \\& = \int_0^1\int_0^1  \frac{x}{1-x^4y^4}\, \mathrm{dy} \, \mathrm{dx} \\& = \frac{1}{2}\int_0^1 (\tan^{-1} x +\tanh^{-1}{x})\,\mathrm{dx} \\& = \frac{1}{8} (π + \log{4}). \end{aligned}
You can establish a far more general formula for $S_n(s)$ but the answer isn't very nice. I believe
$$\displaystyle \displaystyle \sum_{k \ge 0 }\prod_{1 \le r \le j}\frac{1}{(n k+r)}  =  \frac{\sqrt{\pi}}{n(j-1)!}\sum_{0 \le r < j}\binom{j-1}{r}\frac{(-1)^r  \Gamma( \frac{r+1}{n})}{\Gamma(\frac{r+1  }{n}+\frac{1}{2})}$$ achieved by rewriting the LHS as
$$\displaystyle \frac{1}{(j-1)!}\int_{0}^{1}\frac{(1-t)^{j-1}}{1-t^n}\;{dt}$$
via multiple integration or by writing the summand in terms of gamma/beta function.
A: $$\begin{aligned} S&= \sum_{k\geq 0}\frac{1}{(4k+1)(4k+2)} \\& =\sum_{k\geq 0}\left(\frac{1}{4k+1}-\frac{1}{4k+2}\right)\\& =\frac{1}{4}\sum_{k\geq 0} \left(\frac{1}{k+\frac{1}{4}}-\frac{1}{k+\frac{1}{2}}\right)=\frac{1}{4}\left(\psi^0\left(\frac{1}{2}\right)-\psi^0\left(\frac{1}{4}\right)\right)\\& =\frac{1}{4}\left(-\gamma -\ln(4)+\gamma+\frac{\pi}{2}+\ln(8)\right)=\frac{\ln2}{4} +\frac{\pi}{8}\end{aligned} $$
