The consequences of π

Prove that:

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

Solution

We are using the following lemma:

Lemma: It holds that:
$$\sum_{n=2}^{\infty} \frac{2^{n}}{\binom{2 n}{n}} = \frac{\pi}{2}$$

Proof:We begin by the function $\displaystyle f(x)=\frac{\arcsin x}{\sqrt{1-x^2}}$ which has a MacLaurin series given by the formula:

$$f(x) = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle (2 n+1) \binom{2 n}{n}} x^{2 n+1}$$

Differentiating and plugging $x=1/2$ at the derivative yiels the result.

Hence for the series in question we have successively:

$$\begin{aligned}\sum_{n=2}^{\infty} \prod_{k=1}^n \frac{2 k}{n+k-1}&= \sum_{n=2}^{\infty}\frac{2^n n!}{n (n+1) \cdots (2 n-1)}\\ &=\sum_{n=2}^{\infty}\frac{2^n n! (n-1)!}{(2 n-1)!}\\ &= 2 \sum_{n=2}^{\infty}\frac{2^n}{\displaystyle \binom{2 n}{n}} \\&= 2 \cdot \frac{\pi}{2} \\ &= \pi \end{aligned}$$