Trying to prove that $\sum_{j=2}^\infty \prod_{k=1}^j \frac{2 k}{j+k-1} = \pi$ How could one prove that: $$\sum_{j=2}^\infty \prod_{k=1}^j \frac{2 k}{j+k-1} = \pi$$
This is about as far as I got:
$$\prod_{k=1}^j \frac{2 k}{j+k-1} = \frac{2^j j!}{(j)_j} \implies$$
$$\sum_{j=2}^\infty\frac{2^j j!}{(j)_j} = \frac{2^2 2!}{(2)_2} + \frac{2^3 3!}{(3)_3} + \frac{2^4 4!}{(4)_4}+\cdots \implies$$
$$?$$
where $(x)_n$ denotes the Pochhammer symbol.
Maybe reduction isn't the way to go?

This is going to be a little out of the blue, but here goes.
Consider the function
$$f(x) = \frac{\arcsin{x}}{\sqrt{1-x^2}}$$
$f(x)$ has a Maclurin expansion as follows:
$$f(x) = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle (2 n+1) \binom{2 n}{n}} x^{2 n+1}$$
Differentiating, we get
$$f'(x) = \frac{x \, \arcsin{x}}{(1-x^2)^{3/2}} + \frac{1}{1-x^2} = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle \binom{2 n}{n}} x^{2 n}$$
Evaluate at $x=1/\sqrt{2}$:
$$f'\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{2}+2 = \sum_{n=0}^{\infty} \frac{2^{n}}{\displaystyle \binom{2 n}{n}} $$
Thus we have established that
$$\sum_{n=2}^{\infty} \frac{2^{n}}{\displaystyle \binom{2 n}{n}} = \frac{\pi}{2}$$
Now consider the original sum:
$$\begin{align}\sum_{n=2}^{\infty} \prod_{k=1}^n \frac{2 k}{n+k-1}&= \sum_{n=2}^{\infty}\frac{2^n n!}{n (n+1) \cdots (2 n-1)}\\ &=\sum_{n=2}^{\infty}\frac{2^n n! (n-1)!}{(2 n-1)!}\\ &= 2 \sum_{n=2}^{\infty}\frac{2^n}{\displaystyle \binom{2 n}{n}} \\&= 2 \frac{\pi}{2} \\ &= \pi \end{align} $$
QED
