Evaluate: $S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$ Evaluate of this sum:
$$S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$$
Expand out the sum:
$$S=\prod_{k=1}^{1}\frac{2k}{k+2}+\prod_{k=1}^{2}\frac{2k}{k+3}+\prod_{k=1}^{3}\frac{2k}{k+4}+\cdots$$
$$S=\frac{2}{3}+\frac{2}{4}\cdot\frac{4}{5}+\frac{2}{5}\cdot\frac{4}{6}\cdot\frac{6}{7}+
\frac{2}{6}\cdot\frac{4}{7}\cdot\frac{6}{8}\cdot\frac{8}{9}+\cdots+\frac{2^nn!}{(2n)!\div (n+1)!}$$
I don't know what to do next...

Amazing is to recognize some series.
Consider
$$S=\sum_{j=1}^{\infty} \frac{j!\,(j+1)!\, 2^j}{(2j+1)!}x^{2j}$$ Let $x=y \sqrt 2$ to make 
$$S=\sum_{j=1}^{\infty}\frac{4^j\, j!\, (j+1)! }{(2 j+1)!}y^{2 j}=-\frac{y^2}{y^2-1}+\frac{1}{2 \left(y^2-1\right)}-\frac{\sin ^{-1}(y)}{2
   \sqrt{1-y^2} \left(y^2-1\right) y}$$
Making $y=\frac 1 {\sqrt 2}$ gives the result.
If we expand the rhs as a Taylor series built around $y=\frac 1 {\sqrt 2}$ we have
$$S=\frac{\pi }{2}+\sqrt{2} (4+\pi )
   \left(y-\frac{1}{\sqrt{2}}\right)+O\left(\left(y-\frac{1}{\sqrt{2}}\right)^2
\right)$$
$$S=\frac{\pi }{2}+(4+\pi ) (x-1)+O\left((x-1)^2\right)$$
