Prove $\sum _{n \geq 1} \left[\frac{1}{n}-\frac{2}{n+1}{_2F_1}(1,\frac{n+1}{2};\frac{n+3}{2};-1)\right]=\frac{\pi}{4}-\frac{\log 2}{2}$ 
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:
$$I= \int_0^\infty \frac{e^x+e^{-x}-2}{(e^x-1)(e^x+e^{-x})}dx= \\ =\int_0^\infty \frac{dx}{e^x+e^{-x}}-\int_0^\infty \frac{dx}{e^{2x}+1}= \\ =\int_1^\infty \frac{dt}{t^2+1}-\int_1^\infty \frac{dt}{t(t^2+1)}= \frac{\pi}{4}-\frac{\log 2}{2}$$

It's easy enough to work with the hypergeometric function directly, since $(n+1)/2$ and $(n+3)/2$ differ by 1. This means we have
\begin{align}_2F_1\left(1,\frac{n+1}2;\frac{n+3}2;-1\right)&=(n+1)\sum_{k=0}^\infty\frac{(-1)^k}{2k+n+1}\\&=(n+1)\sum_{k=0}^\infty\int_0^1(-x^2)^kx^n~\mathrm dx\\&=(n+1)\int_0^1\sum_{k=0}^\infty(-x^2)^kx^n~\mathrm dx\\&=(n+1)\int_0^1\frac{x^n}{1+x^2}~\mathrm dx\end{align}
Shoving this into your sum gives
\begin{align}S&=\sum_{n=1}^\infty\frac1n-2\int_0^1\frac{x^n}{1+x^2}~\mathrm dx\\&=\sum_{n=1}^\infty\int_0^1x^{n-1}-\frac{2x^n}{1+x^2}~\mathrm dx\\&=\int_0^1\sum_{n=1}^\infty x^{n-1}-\frac{2x^n}{1+x^2}~\mathrm dx\\&=\int_0^1\frac1{1-x}-\frac{2x}{(1+x^2)(1-x)}~\mathrm dx\end{align}
which is easy to integrate with partial fractions and gives the result smoothly.
Hopefully this is what you wanted.
