Euler's work VI

Series related to the Catalan constant

We show te proof of the following result. \[ \sum_{n=0}^{\infty} \frac{(2n)!!(n+1)(2n+5)}{(2n+1)!!(2n+1)(2n+3)^2}=1\] The proof involves one of the series representation of the Catalan constant.

Proof

First note that \[ \frac{(n+1)(2n+5)}{(2n+1)(2n+3)^2} = \frac{1}{2(2n+1)}+\frac{1}{2(2n+3)^2}\] Hence \begin{align*} S=\sum_{n=0}^{\infty} \frac{(2n)!!(n+1)(2n+5)}{(2n+1)!!(2n+1)(2n+3)^2} =& \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!}\left[\frac{1}{2(2n+1)}+\frac{1}{2(2n+3)^2}\right] \\ =& \frac{1}{2}\underbrace{ \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!(2n+1)}}_{S_{1}} + \frac{1}{2}\underbrace{\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!(2n+3)^2}}_{S_{2}} \end{align*} In a previous post we evaluated $S_{1}$: \[ S_{1}= \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!(2n+1)} = 2\beta(2) \] where $\beta(2)$ is the Calatan constant.

For $S_{2}$:

\begin{align*} S_{2} =& \sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!(2n+3)^2}\\ =&\sum_{n=0}^{\infty}\frac{(2n+2)!!}{(2n+3)!!(2n+3)(2n+2)}\\ =&\sum_{j=1}^{\infty}\frac{(2j)!!}{(2j+1)!!(2j+1)(2j)} \quad (n=j-1) \end{align*} Note that \[\underbrace{\sum_{n=1}^{\infty} \frac{(2j)!!}{(2j+1)!!(2j+1)}}_{2\beta(2)-1} = \underbrace{\sum_{j=1}^{\infty}\frac{(2j)!!}{(2j+1)!!2j}}_{S_{3}} - \underbrace{\sum_{j=1}^{\infty}\frac{(2j)!!}{(2j+1)!!(2j+1)(2j)}}_{S_{2}}\] We just have to find $S_{3}$:

In another post we found the following series expansion: \[ \frac{\arcsin(x)}{\sqrt{1-x^2}} = \sum_{j=0}^{\infty} \frac{(2j)!!}{(2j+1)!!}x^{2j+1}\] Dividing by $x^2$ \[ \frac{\arcsin(x)}{x^2\sqrt{1-x^2}} = \sum_{j=0}^{\infty} \frac{(2j)!!}{(2j+1)!!}x^{2j-1} = \frac{1}{x} + \sum_{j=1}^{\infty} \frac{(2j)!!}{(2j+1)!!}x^{2j-1} \] \[ \Longrightarrow \sum_{j=1}^{\infty} \frac{(2j)!!}{(2j+1)!!}x^{2j-1} = \frac{\arcsin(x)}{x^2\sqrt{1-x^2}}- \frac{1}{x} \] Integrating from $0$ to $1$: \[ \Longrightarrow S_{3} = \sum_{j=1}^{\infty} \frac{(2j)!!}{(2j+1)!!2j} = \int_{0}^{1}\frac{\arcsin(x)}{x^2\sqrt{1-x^2}}- \frac{1}{x} dx \] \[ I= \int \frac{\arcsin(x)}{x^2\sqrt{1-x^2}}- \frac{1}{x} dx = \int\frac{\arcsin(x)}{x^2\sqrt{1-x^2}}dx - \ln(x) + C\] \begin{align*} \int\frac{\arcsin(x)}{x^2\sqrt{1-x^2}}dx =& \int \frac{w}{\sin^2(w)}dw \quad (w \mapsto \arcsin(x))\\ \stackrel{IBP}{=}& -w\cot(w) + \int \cot(w) dw \\ =& -w\cot(w) +\ln(\sin w) + C\\ =& -\frac{\sqrt{1-x^2}\arcsin(x)}{x} +\ln(x) + C \end{align*} Therefore \[ I = \int \frac{\arcsin(x)}{x^2\sqrt{1-x^2}}- \frac{1}{x} dx = -\frac{\sqrt{1-x^2}\arcsin(x)}{x}+ C\] Hence \[ S_{3} = \int_{0}^{1}\frac{\arcsin(x)}{x^2\sqrt{1-x^2}}- \frac{1}{x} dx = \left[-\frac{\sqrt{1-x^2}\arcsin(x)}{x}\right]_{0}^{1} = 1\] Therefore \[S_{2} = S_{3} -\beta(2) +1 = 2-2\beta(2) \] \[S = \frac{1}{2}S_{1} + \frac{1}{2}S_{2} = \beta(2) + 1 - \beta(2) = 1\] We can conclude: \[ \boxed{\sum_{n=0}^{\infty} \frac{(2n)!!(n+1)(2n+5)}{(2n+1)!!(2n+1)(2n+3)^2}=1}\]