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...
 A: I think, your expansion of the product is incorrect.
I get:
$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}=\sum_{j=1}^{\infty}\frac{2^j j!}{\frac{(2j+1)!}{(j+1)!}}=\sum_{j=1}^{\infty} 2^j \frac{j! (j+1)!}{(2j+1)!} =\sum_{j=1}^{\infty} \frac{2^j}{\binom{2j+1}{j}} = \frac{\pi}{2}$$
Edit:
If you want to know how to get to $\frac{\pi}{2}$ then look at the question and answer from: How to sum this series for π/2
directly?
A: Correcting $(2n)!$ to $(2n+1)!$ in OP's nth term, we have
$$S=\sum_{n=1}^{\infty} \frac{2^n n! (n+1)!}{(2n+1)!} =\sum_{n=1}^{\infty}\frac{2^n n ~\Gamma(n) \Gamma(n+2)}{\Gamma(2n+2)}=\sum_{n=1}^{\infty}2^n n~B(n,n+2)= \sum_{n=1}^{\infty} 2n\int_{0}^{\pi/2} 2^n \sin^{2n-1} x ~\cos^{2n+3} xdx.$$
$$S=\int_{0}^{\pi/2} \sqrt{2}\cos^4 x \sum_{n=1}^{\infty} n~(\sqrt{2} \sin x \cos x)^{2n-1}$$
Using the infinite GP result that $\sum_{1}^{\infty} n ~z^{2n-1} =\frac{z}{(1-z^2)^2}$ We get
$$S=\int_{0}^{\pi/2}  \sqrt{2} \cos^4 x \frac{\sqrt{2} \sin x \cos x}{(1-2\sin^2 x\cos^2 x)^2} dx=
\int_{0}^{\pi/2} \frac{2 \sin x \cos^5x}{(1-2\sin^2 x \cos^2 x)^2} dx$$
Next using $\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx,$
we get $$S=\int_{0}^{\pi/2} \frac{2 \sin^5 x \cos x}{(1-2\sin^2 x \cos^2 x)^2} dx$$ Adding the last two integrals, we get
So $$2S=\int_{0}^{\pi/2} \frac{2\sin x \cos x(\sin^4 x+ \cos^4 x)}{(1-2\sin^2 x \cos^2 x)^2}=\int_{0}^{\pi/2} \frac{2\sin x \cos x}{(1-2\sin^2 x \cos^2 x)}dx=\int_{0}^{\pi/2} \frac{4\sin 2x dx}{1+\cos^2 2x}$$
$$\implies 2S=8\int_{0}^{\pi/4}\frac{\sin 2x dx}{1+\cos^2 2x}=-4\int_{1}^{0}\frac{dt}{1+t^2}=\pi \implies S=\frac{\pi}{2}$$
Lastly we have used $\cos 2x=t.$ 
A: The product can be simplified to:
$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}=
\sum_{j=1}^{\infty} \frac{j!(j+1)! 2^j}{(2j+1)!}=
\sum_{j=1}^{\infty} \frac{2^j}{\binom{2j+1}{j}}$$
A: 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
$$S=\frac{\pi }{2}+\sqrt{2} (4+\pi )
   \left(y-\frac{1}{\sqrt{2}}\right)+O\left(\left(y-\frac{1}{\sqrt{2}}\right)^2
\right)$$
$$S=\frac{\pi }{2}+(4+\pi ) (x-1)+O\left((x-1)^2\right)$$
