Hypergeometric, Seriously?

$\displaystyle _2F_1(a,b;c;z) = \dfrac{\Gamma{(c)}}{\Gamma{(b)}\Gamma{(c-b)} } \int_0^1 x^{b-1} (1-x)^{c-b-1} (1-zx)^{-a} dx$

Matching with this ,one can tell that

$\displaystyle\sum_{n=1}^∞ \dfrac{_2F_1(-n,\frac{1}{4};\frac{3}{4};2)}{n} = \dfrac{\Gamma{(\frac{3}{4})}}{\Gamma{(\frac{1}{2})}\Gamma{(\frac{1}{4})}}\int_0^1 x^{-\frac{3}{4}} (1-x)^{-\frac{1}{2}} \sum_{n=1}^∞ \dfrac{(1-2x)^n}{n} dx$ $\displaystyle=-\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{-\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}-{x}\right)^{-\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{log}\:\left(\mathrm{2}{x}\right){dx} =\frac{-\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\mathrm{log}\:\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{-\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}-{x}\right)^{-\frac{\mathrm{1}}{\mathrm{2}}} {dx}-\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{-\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}-{x}\right)^{-\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{log}\left({x}\right){dx}$

$\displaystyle=-\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\mathrm{log}\left(\mathrm{2}\right)\boldsymbol{\mathrm{B}}\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{2}}\right)-\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}.\frac{\partial}{\partial{a}}\mid_{{a}=\frac{\mathrm{1}}{\mathrm{4}}} \frac{\Gamma\left({a}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)}$

$\displaystyle=-\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\mathrm{log}\:\left(\mathrm{2}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}.\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}-\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}.\left(\frac{\Gamma'\left({a}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)}-\frac{\Gamma'\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({a}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma^{\mathrm{2}} \left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\mid_{{a}=\frac{\mathrm{1}}{\mathrm{4}}} \right)$

$\displaystyle=-\mathrm{log}\:\left(\mathrm{2}\right)-\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\left(\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)-\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\right)$

$\displaystyle=-\mathrm{log}\:\left(\mathrm{2}\right)-\left(\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)-\psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right)$

Now , $\displaystyle \psi(1-a)-\psi(a)= \pi \cot(\pi a)$ And $\displaystyle\psi(\frac{3}{4}) -\psi(\frac{1}{4}) = \pi$

So answer is $\boxed{\pi-\log(2)= 2.44}$

Relevant wiki

• Digamma function

• Gamma function