Euler's work

Euler and Catalan constant

We show the proof of this interesting series related to the work of Euler and the Catalan constant. \[ \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!(2n+1)} = 2\beta(2) \]
Proof
From a previous post (link) we found that \[\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} dx = 2\beta(2) \] Therefore \[2\beta(2) = \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} dx =\int_{0}^{1} \frac{\arcsin w}{w \sqrt{1-w^2}} dw \quad (w \mapsto \sin x) \tag{1} \] Now we will show that \[\frac{\arcsin w}{\sqrt{1-w^2}} = \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} w ^{2n+1} \] Euler found the following formula for $\arctan(x)$: \[\arctan(x) = \frac{x}{1+x^2}\left[1+\frac{2}{3}\frac{x^2}{1+x^2}+\frac{2}{3}\frac{4}{5}\left(\frac{x^2}{1+x^2}\right)^2+\frac{2}{3}\frac{4}{5}\frac{6}{7}\left(\frac{x^2}{1+x^2}\right)^3+...\right]=\frac{x}{1+x^2}\sum_{n=0}^{\infty}\prod_{k=1}^{\infty} \frac{2kx^2}{(2k+1)(1+x^2)} \tag{2}\] Now, consider the following identity for inverse trigonometric functions: \[\arcsin(w) = \arctan\left(\frac{w}{\sqrt{1-w^2}}\right)\] Then if we substitute $x= \displaystyle \frac{w}{\sqrt{1-w^2}}$ in (2) we have \begin{align*} \arcsin(w) =& \arctan\left(\frac{w}{\sqrt{1-w^2}}\right) = w\sqrt{1-w^2}\left[1+\frac{2}{3}w^2 +\frac{2}{3}\frac{4}{5}w^4 + \frac{2}{3}\frac{4}{5}\frac{6}{7}w^6...\right]\\ \Longrightarrow \frac{\arcsin(w)}{\sqrt{1-w^2} }=& w+\frac{2}{3}w^3+\frac{2}{3}\frac{4}{5}w^5+\frac{2}{3}\frac{4}{5}\frac{6}{7}w^7+...\\ =& \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} w ^{2n+1} \tag{3} \end{align*} Hence, from (1) and (3) \begin{align*} \int_{0}^{1} \frac{\arcsin w}{w \sqrt{1-w^2}} dw =& \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} \frac{w ^{2n+1}}{w} dw \\ =& \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} w^{2n} dw\\ =& \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} \int_{0}^{1} w^{2n} dw\\ =& \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!(2n+1)} \end{align*} Therefore \[ \boxed{\sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!(2n+1)} = 2\beta(2)} \]