SUMS OF RECIPROCALS OF THE CENTRAL BINOMIAL COEFFICIENTS

Today we compute sums of reciprocals of the central Binomial Coefficients, namely

$\begin{aligned} &\sum_{n=1}^{\infty} \frac{1}{n^{2}\left(\begin{array}{c} 2 n \\ n \end{array}\right)}=\frac{\pi^{2}}{18} \\ &\sum_{n=1}^{\infty} \frac{2^{n}}{n^{2}\left(\begin{array}{c} 2 n \\ n \end{array}\right)}=\frac{\pi^{2}}{8} \\ &\sum_{n=1}^{\infty} \frac{3^{n}}{n^{2}\left(\begin{array}{c} 2 n \\ n \end{array}\right)}=\frac{2 \pi^{2}}{9} \end{aligned}$

As we will see, it´s intimately related to the series expansion of the arcsine function.

First recall the trigonometric form of the Beta function (see this post)

$\int_0^{\pi/2}\left(\sin\left(t\right)\right)^{2x-1}\left(\cos\left(t\right)\right)^{2y-1}\,dt=\frac12\frac{\Gamma\left(x\right)\Gamma\left(y\right)}{\Gamma\left(x+y\right)}=B(x,y)$

if we let $y=\frac12$ we obtain

$\int_0^{\pi/2}\left(\sin\left(t\right)\right)^{2n-1}\,dt=\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{1}{2n}$ (1)

Proof:

$\begin{aligned} \int_0^{\pi/2}\left(\sin\left(t\right)\right)^{2n-1}\,dt&=\frac12\frac{\Gamma\left(n\right)\Gamma\left(\frac12\right)}{\Gamma\left(n+\frac12\right)}\\ &=\frac12\frac{\Gamma\left(n\right)\sqrt{\pi}}{\Gamma\left(n+\frac12\right)}\\ &=\frac12\frac{\Gamma\left(n\right)\sqrt{\pi}}{\frac{\Gamma\left(2n\right)\sqrt{\pi}}{2^{2n-1}\Gamma(n)}} \qquad (*)\\ &=\frac{2^{2n}\Gamma^2\left(n\right)}{4\Gamma\left(2n\right)}\\ &=\frac{4^{n-1}}{\frac{\Gamma\left(2n\right)}{\Gamma^2\left(n\right)}}\\ &=\frac{4^{n-1}}{\frac{2n!}{2n}\frac{n^2}{n!^2}}\\ &=\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{1}{2n} \qquad \blacksquare\\ \end{aligned}$

Where in (*) we made use of Legendre´s duplication formula proved here

$\Gamma(2 x)=\frac{2^{2 x-1} \Gamma(x) \Gamma\left(x+\frac{1}{2}\right)}{\sqrt{\pi}}$

Now we multiply (1) by $x^{2n-1}$ and sum from 1 to $\infty$ we get

$\sum_{n=1}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n-1}}{2n}=\frac{\arcsin(x)}{\sqrt{1-x^2}}$ (2)

Proof:

$\begin{aligned} \frac12 \sum_{n=1}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n-1}}{n}&=\sum_{n=1}^\infty\int_0^{\pi/2}\left(x\sin\left(t\right)\right)^{2n-1}\,dt\\ &=\sum_{n=1}^\infty\int_0^{\pi/2}\frac{\left(x\sin\left(t\right)\right)^{2n}}{x\sin\left(t\right)}\,dt\\ &=\int_0^{\pi/2}\frac{1}{x\sin\left(t\right)}\sum_{n=1}^\infty\left(x^2\sin^2\left(t\right)\right)^{n}\,dt\\ &=\int_0^{\pi/2}\frac{1}{x\sin\left(t\right)}\frac{x^2\sin^2\left(t\right)}{1-x^2\sin^2\left(t\right)}\,dt\\ &=\int_0^{\pi/2}\frac{x\sin\left(t\right)}{1-x^2\sin^2\left(t\right)}\,dt\\ &=\int_0^{\pi/2}\frac{x\sin\left(t\right)}{\cos^2(t)+\sin^2(t)-x^2\sin^2\left(t\right)}\,dt\\ &=\int_0^{\pi/2}\frac{x\sin\left(t\right)}{\cos^2(t)+\left(1-x^2\right)\sin^2\left(t\right)}\,dt\\ &=\int_0^{\pi/2}\frac{x\sin\left(t\right)}{\cos^2(t)+\left(1-x^2\right)\left(1-\cos^2\left(t\right)\right)}\,dt\\ &=\int_0^{\pi/2}\frac{x\sin\left(t\right)}{1-x^2+x^2\cos^2(t)}\,dt\\ &=x\int_0^{1}\frac{1}{1-x^2+x^2t^2}\,dt \qquad \left( \cos(t) \to t\right)\\ &=\frac{x}{1-x^2}\int_0^{1}\frac{1}{1+\frac{x^2t^2}{1-x^2}}\,dt\\ &=\frac{1}{\sqrt{1-x^2}}\int_0^{\frac{x}{\sqrt{1-x^2}}}\frac{1}{1+t^2}\,dt \qquad \left(t \to \frac{\sqrt{1-x^2}}{x}t \right)\\ &=\frac{1}{\sqrt{1-x^2}}\arctan\left(\frac{x}{\sqrt{1-x^2}}\right) \\ &=\frac{\arcsin(x)}{\sqrt{1-x^2}} \qquad \blacksquare \end{aligned}$

Now we integrate (2) to obtain

$\arcsin^2(x)=\frac12\sum_{n=1}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n}}{n^2}$ (3)

Proof:

First we show the L.H.S. of (3)

$\begin{aligned} \int\frac{\arcsin(x)}{\sqrt{1-x^2}} \,dx&=\int x \,dx \qquad \left(x \to \arcsin(x) \right)\\ &=\frac{x^2}{2}+C\\ &=\frac{\arcsin^2(x)}{2}+C \end{aligned}$

For the R.H.S.

$\begin{aligned} \frac12\sum_{n=1}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{1}{n}\int x^{2n-1}\,dx&=\frac14\sum_{n=1}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{1}{n^2}\\ \end{aligned}$

Therefore

$\begin{aligned} \frac14\sum_{n=1}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{1}{n^2}&=\frac{\arcsin^2(x)}{2}+C\\ \end{aligned}$

Letting x=0 we find that C=0 and finally

$2\arcsin^2(x)=\sum_{n=1}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n}}{n^2} \qquad \blacksquare$

If we now let $x=\frac12$ in (3) we get

$\begin{aligned} \sum_{n=1}^\infty\frac{ 1}{n^2\left(\begin{array}{c}2n\\ n\end{array}\right)} &=2\arcsin^2\left(\frac12\right)\\ &=2\left(\frac{\pi}6\right)^2\\ &=\frac{\pi^2}{18} \qquad \blacksquare\\ \end{aligned}$

Setting $x=\sqrt{\frac12}$ in (3)

$\begin{aligned} \sum_{n=1}^\infty\frac{ 2^n}{n^2\left(\begin{array}{c}2n\\ n\end{array}\right)} &=2\arcsin^2\left(\sqrt{\frac12}\right)\\ &=2\left(\frac{\pi}4\right)^2\\ &=\frac{\pi^2}{8} \qquad \blacksquare\\ \end{aligned}$

Finally letting $x=\frac{\sqrt{3}}{2}$ in (3) we obtain

$\begin{aligned} \sum_{n=1}^\infty\frac{ 3^n}{n^2\left(\begin{array}{c}2n\\ n\end{array}\right)} &=2\arcsin^2\left(\frac{\sqrt{3}}{2}\right)\\ &=2\left(\frac{\pi}3\right)^2\\ &=\frac{2\pi^2}{9} \qquad \blacksquare\\ \end{aligned}$

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