<s>[INST] 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...
 [/INST] 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 $\cos 2x=t.$ 
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