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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?

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    $\begingroup$ The Pochhammer symbol is a notation, not an operation; the operation here is the rising factorial. (I only mention it because in many fields, the Pochhammer symbol is used to denote the falling factorial instead.) $\endgroup$ – Nick Peterson Jun 15 '13 at 3:06
  • $\begingroup$ I'd be interested to see if there was a complex analytic proof. $\endgroup$ – Alexander Gruber Jun 15 '13 at 3:53
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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

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    $\begingroup$ How did you pull that out of the blue?! +1 $\endgroup$ – JohnWO Jun 15 '13 at 10:03
  • $\begingroup$ @JohnWO: working backwards. I computed a generating function for the terms in the series you posted and massaged it until I got a simple and presentable result. $\endgroup$ – Ron Gordon Jun 15 '13 at 10:05
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    $\begingroup$ It's a bit of a tour de force but it's quite impressive and enlightening. $\endgroup$ – Cameron Williams Jun 21 '14 at 23:41
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    $\begingroup$ I just gotta say... I just re-reviewed this, and it amazes me to this day that you pulled this out of the blue! $\endgroup$ – JohnWO Jul 4 '15 at 2:09
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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 $$\sum_{j=2}^\infty\dfrac{2^j j!}{(j)_j} = \pi$$

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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

$$ 4 + \pi = \sum_{n=0}^{\infty} (-1)^n a_n = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(\Delta^n a)_0 = \sum_{n=0}^{\infty}b_n=2 + 2 + \sum_{n=2}^{\infty}b_n. $$

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{j = 2}^{\infty}\prod_{k = 1}^{j}{2k \over j + k - 1} = \pi:\ {\large ?}}$

\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.

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