I was trying to prove the problem here problem no 17
$$\displaystyle \sum_{n=0}^{\infty} \dfrac{2^{n-1} n!}{\prod_{r=0}^{n} (4r+1)} = \dfrac{\pi + 2\log(1+\sqrt{2})}{4 \sqrt{2}} $$
My Try: $$\sum_{n=0}^{\infty} \dfrac{2^{n-1} n!}{\prod_{r=0}^{n} (4r+1)} =\frac{\Gamma(5/4)}{\Gamma(3/4)}\sum_{n=0}^{\infty}2^{-n-1}B(n+1,3/4)$$ $$=\frac{\Gamma(5/4)}{\Gamma(3/4)}\sum_{n=0}^{\infty}2^{-n-1}\sum_{k \geq 0}\frac{\binom{k-3/4}{k}}{n+k+1}$$ $$=\frac12 \frac{\Gamma(5/4)}{\Gamma(3/4)}\sum_{k \geq 0}\frac{\binom{k-3/4}{k}}{n+k+1}\Phi(\frac12,1,k+1)$$ $$=\frac{\Gamma(5/4)}{\Gamma(3/4)}\int\limits_{0}^{\infty}\frac{e^{t/4}}{(e^t-1)^{1/4}(2e^t-1)}dt$$ Now stuck at this point, can anyone help??
I prefer to re-start from scratch: $$\begin{eqnarray*}\sum_{n\geq 0}\frac{2^{n-1}n!}{\prod_{r=0}^{n}(4r+1)}=\sum_{n\geq 0}\frac{2^{n-1}\Gamma(n+1)\Gamma\left(\frac{5}{4}\right)}{4^n \Gamma\left(\frac{5}{4}+n\right)}&=&\frac{1}{2}+\sum_{n\geq 1}n2^{-(n+1)}B\left(n,\frac{5}{4}\right)\\&=&\frac{1}{2}+\int_{0}^{1}\sum_{n\geq 1}n 2^{-(n+1)} t^{n-1}(1-t)^{1/4}\,dt\\&=&\frac{1}{2}+\int_{0}^{1}\frac{(1-t)^{1/4}}{(2-t)^2}\,dt\\&=&\frac{1}{2}+\int_{0}^{1}\frac{s^{1/4}}{(1+s)^2}\,ds\\&=&\frac{1}{2}+4\int_{0}^{1}\frac{u^4}{(1+u^4)^2}\,du\end{eqnarray*}$$ and the last integral can be computed from partial fraction decomposition, through: $$ (u^4+1)=(u^2+u\sqrt{2}+1)(u^2-u\sqrt{2}+1).$$