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Can we prove the following closed form by directly using hypergeometric series? $$S=\sum _{n=1}^{\infty} \left[\frac{1}{n}-\frac{2 }{n+1}\, _2F_1\left(1,\frac{n+1}{2};\frac{n+3}{2};-1\right)\right]=\frac{\pi}{4}-\frac{\log 2}{2}$$

I have come to this result by considering the integral:

$$I = \int_0^\infty \frac{1-\frac{1}{\cosh x}}{e^x-1} dx$$

We can expand the denominator as geometric series and then find the integrals for each term.

$$\int_0^\infty \frac{2 e^{-n x}dx}{e^x+e^{-x}}=2 \int_0^\infty \frac{t^n dt}{1+t^2}=\int_0^\infty \frac{u^{(n-1)/2} du}{1+u}= \\ = B \left(1,\frac{n+1}{2} \right) \, _2F_1\left(1,\frac{n+1}{2};\frac{n+3}{2};-1\right)$$

As for the integral as a whole, it's easy enough to rewrite it:

$$I= \int_0^\infty \frac{e^x+e^{-x}-2}{(e^x-1)(e^x+e^{-x})}dx= \\ =\int_0^\infty \frac{dx}{e^x+e^{-x}}-\int_0^\infty \frac{dx}{e^{2x}+1}= \\ =\int_1^\infty \frac{dt}{t^2+1}-\int_1^\infty \frac{dt}{t(t^2+1)}= \frac{\pi}{4}-\frac{\log 2}{2}$$