Trying to prove that $\sum_{j=2}^\infty \prod_{k=1}^j \frac{2 k}{j+k-1} = \pi$

Question

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?

Answer 1

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

Answer 2

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 $$\sum_{j=2}^\infty\dfrac{2^j j!}{(j)_j} = \pi$$

Answer 3

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

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