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?
 A: First of all you can use Mathematica to compute it and, well, get the same result. :)
But anyways, we need an analytic solution. The main idea is to put it into the form of hypergeometric series.
I'll catch where you stopped:
$$\prod_{k=1}^j \dfrac{2 k}{j+k-1} = \dfrac{2^j j!}{(j)_j}$$
One can reorganize it like:
$$\dfrac{2^j j!}{(j)_j}=\dfrac{2^j \Gamma (j) \Gamma (j+1)}{\Gamma (2 j)}$$
because $(j)_j=\dfrac{\Gamma (2 j)}{\Gamma (j) }$.
Using the fact that $$\Gamma (2 j)=\dfrac{2^{2j-1}}{\sqrt{\pi}}\Gamma(j)\Gamma\left(j+\dfrac{1}{2}\right)$$
one can obtain:
$$\dfrac{2^j j!}{(j)_j}=2\sqrt{\pi}\dfrac{\Gamma (j+1)}{2^j\Gamma\left(j+\dfrac{1}{2}\right)}=2\sqrt{\pi}\dfrac{j!}{\Gamma\left(j+\dfrac{1}{2}\right)}\left(\dfrac{1}{2}\right)^j$$
Then one can rewrite it in terms of the Pochhammer symbols. Keeping in mind that $\Gamma\left(j+\dfrac{1}{2}\right)=\sqrt{\pi}\left(\dfrac{1}{2}\right)_j$ and $j!=(1)_j$ and after multiplying and deviding by $j!=(1)_j$ one will obtain:
$$\dfrac{2^j j!}{(j)_j}=2\dfrac{(1)_j(1)_j}{\left(\dfrac{1}{2}\right)_j}\left(\dfrac{1}{2}\right)^j$$
So, the initial series will look like:
$$\sum_{j=2}^\infty\frac{2^j j!}{(j)_j} =\sum_{j=0}^\infty\frac{2^j j!}{(j)_j}-2 =2\sum_{j=0}^\infty\dfrac{(1)_j(1)_j}{\left(\dfrac{1}{2}\right)_j}\dfrac{\left(\dfrac{1}{2}\right)^j}{j!}-4$$
And the first term is the definition of the hypergeometric function $$\sum_{j=0}^\infty\dfrac{(1)_j(1)_j}{\left(\dfrac{1}{2}\right)_j}\dfrac{\left(\dfrac{1}{2}\right)^j}{j!}=\, _2F_1\left(1,1;\dfrac{1}{2};\dfrac{1}{2}\right)=\dfrac{4+\pi }{2}$$
So $$\sum_{j=2}^\infty\dfrac{2^j j!}{(j)_j} = \pi$$
A: 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
A: Define the sequence $b$ by $b_0 = 2$ and $$b_n = \prod_{k=1}^n\frac{2k}{n+k-1}$$ for $n \geq 1$, so we want to compute $\sum_{n=2}^{\infty}b_n$.
Let $a_0, a_1, a_2, \dotsc$ be the sequence $$4, -4, -\frac{4}{3}, -\frac{4}{5}, -\frac{4}{7}, \dotsc$$
and let $\Delta$ be the difference operator on sequences defined by $(\Delta \alpha)_n = \alpha_{n+1} -\alpha_n$.  Then one can check that the repeated difference at $0$ equals
$$(\Delta^n a)_0 = (-1)^n2^{n+1} b_n.
$$
By Euler's transform we get
$$
4 + \pi = \sum_{n=0}^{\infty} (-1)^n a_n = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(\Delta^n a)_0 = \sum_{n=0}^{\infty}b_n=2 + 2 + \sum_{n=2}^{\infty}b_n.
$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{j = 2}^{\infty}\prod_{k = 1}^{j}{2k \over j + k - 1} = \pi:\ {\large ?}}$

\begin{align}
&\color{#c00000}{\sum_{j = 2}^{\infty}\prod_{k = 1}^{j}{2k \over j + k - 1}}
=\sum_{j = 2}^{\infty}2^{j}\,
{1 \over j}\,{2 \over j + 1}\cdots{j \over 2j - 1}=\sum_{j = 2}^{\infty}2^{j}\,{j! \over \pars{2j - 1}!/\pars{j - 1}!}
\\[3mm]&=\sum_{j = 2}^{\infty}2^{j}\,
{\Gamma\pars{j + 1}\Gamma\pars{j} \over \Gamma\pars{2j}}
=\sum_{j = 2}^{\infty}2^{j}\,j
{\Gamma\pars{j}\Gamma\pars{j} \over \Gamma\pars{2j}}
=\sum_{j = 2}^{\infty}2^{j}\,j\,{\rm B}\pars{j,j}
\\[3mm]&=2\lim_{x \to 2}\bracks{%
\partiald{}{x}\color{#00f}{\sum_{j = 2}^{\infty}x^{j}\,{\rm B}\pars{j,j}}}
\end{align}
  where $\ds{\Gamma\pars{z}}$ and
  $\ds{{\rm B}\pars{x,y} = \int_{0}^{1}t^{x - 1}\pars{1 - t}^{y - 1}\,\dd t = {\Gamma\pars{x}\Gamma\pars{y} \over \Gamma\pars{x + y}}}$ 
  are the
  Gamma and
  Beta Functions, respectively. $\ds{\Re\pars{x} > 0, \Re\pars{y} > 0}$. We used the property
  $\ds{\Gamma\pars{z} = \Gamma\pars{z + 1}/z}$

\begin{align}
&\color{#00f}{\sum_{j = 2}^{\infty}x^{j}\,{\rm B}\pars{j,j}}=
\sum_{j = 2}^{\infty}x^{j}\,\int_{0}^{1}t^{j - 1}\pars{1 - t}^{j - 1}\,\dd t
=\int_{0}^{1}\sum_{j = 2}^{\infty}\bracks{xt\pars{ 1- t}}^{j}
\,{\dd t \over t\pars{1 - t}}
\\[3mm]&=\int_{0}^{1}{\bracks{xt\pars{ 1- t}}^{2} \over 1 - xt\pars{1 - t}}
\,{\dd t \over t\pars{1 - t}}
=\int_{0}^{1}{x^{2}t\pars{ 1- t} \over 1 - xt\pars{1 - t}}\,\dd t
\end{align}

\begin{align}
&\color{#c00000}{\sum_{j = 2}^{\infty}\prod_{k = 1}^{j}{2k \over j + k - 1}}
=2\
\overbrace{\int_{0}^{1}\bracks{-1 + {1 \over \bracks{1 - 2\pars{1 - t}t}^{2}}}
\,\dd t}^{\ds{=\ {\pi \over 2}}} = \color{#00f}{\Large\pi}
\end{align}

The last integral is trivially evaluated by 'completing the square' in the denominator.
