CENTRAL BINOMIAL AND CATALAN´S CONSTANT

Let´s proof the following two infinite sums related to Catalan´s constant


\sum_{n=0}^\infty\frac{ 4^{n}}{\left(2n+1\right)^2\left(\begin{array}{c}2n\\ n\end{array}\right)}=2\beta(2)


\sum_{n=0}^\infty\frac{1}{(2n+1)^2\binom{2n}{n}}=\frac83\beta(2)-\frac{\pi}3\ln(2+\sqrt{3})


Where \beta(2) is the Catalan´s constant

Recall (see here)


\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\frac{4^n x^{2n+1}}{(2n+1)\binom{2n}{n}}(1)


Dividing both sides by x and integrating from 0 to 1


\sum_{n=0}^\infty\frac{ 4^{n}}{\left(2n+1\right)^2\binom{2n}{n}}=\int_0^1\frac{\arcsin(x)}{x\sqrt{1-x^2}}\,dx(2)

Than


\begin{aligned} \int_0^1\frac{\arcsin(x)}{x\sqrt{1-x^2}}\,dx&=\int_0^{\pi/2}\frac{x \cos(x)}{\sin(x)\sqrt{1-\sin^2(x)}}\,dx\\ &=\int_{0}^{\pi / 2} \frac{x}{\sin (x)} d x\\&=\frac{1}{2} \int_{0}^{\pi / 2} \frac{x}{\cos \left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right)} d x \\ &=\frac{1}{2} \int_{0}^{\pi / 2} \frac{x}{\cos \left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right)} \frac{1}{\frac{\cos \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}} d x \\ &=\frac{1}{2} \int_{0}^{\pi / 2} \frac{x}{\cos ^{2}\left(\frac{x}{2}\right) \tan \left(\frac{x}{2}\right)} d x\\ &=\frac{1}{2} \int_{0}^{\pi / 2} \frac{x \sec ^{2}\left(\frac{x}{2}\right)}{\tan \left(\frac{x}{2}\right)} d x\\ &=2 \int_{0}^{1} \frac{\arctan (x)}{x} d x \qquad\left(\tan \left(\frac{x}{2}\right) \to x \right)\\ &=2 \int_{0}^{1} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)} x^{2 k} d x \\ &=2 \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2}}\\ &=2 \beta(2) \qquad \blacksquare \end{aligned} \\


Dividing both sides of (1) by x and integrating from 0 to 1/2 we obtain


\int_0^{1/2}\frac{\arcsin(x)}{x\sqrt{1-x^2}}\,dx=\frac12\sum_{n=0}^\infty\frac{ 1}{(2n+1)^2\binom{2n}{n}}(3)


Than


\begin{aligned} \sum_{n=0}^\infty\frac{ 1}{(2n+1)^2\binom{2n}{n}}&=2\int_0^{1/2}\frac{\arcsin(x)}{x\sqrt{1-x^2}}\,dx\\ &=2\int_0^{\pi/6}\frac{x}{\sin(x)}\,dx &(x \to \sin(x))\\ &=2\left(\frac43\beta(2)-\frac{\pi}{6}\ln\left(2+\sqrt{3} \right) \right)\\ &=\frac83\beta(2)-\frac{\pi}{3}\ln\left(2+\sqrt{3} \right) \qquad \blacksquare \end{aligned}


Where we used the result proved here:


\begin{aligned} & \int_0^{\pi/6}\frac{x}{\sin\left(x\right)}\,dx=\frac43\beta(2)-\frac{\pi}{6}\ln\left(2+\sqrt{3} \right) \end{aligned}

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