# Sum the infinite series

How to solve this:

\begin{equation*} \sum_{n=1}^{\infty }\left[ \frac{1\cdot 3\cdot 5\cdots \left( 2n-1\right) }{ 2\cdot 4\cdot 6\cdots 2n}\right] ^{3} \end{equation*}

I can make the bracket thing, $\left[ C(2n,n)/4^{n}\right] ^{3}$, but how to proceed now.

• For some basic information about writing math at this site see e.g. here, here, here and here. – Américo Tavares Apr 11 '15 at 12:41
• Let $\quad F(k)~=~\displaystyle\sum_{n=0}^\infty{2n\choose n}^k~x^n.\quad$ Then $\quad F(0)~=~\dfrac1{1-x}~,\quad F(1)~=~\dfrac1{\sqrt{1-4x}}~,\quad F(2)~=$ $=~\dfrac2\pi~K\big(4~\sqrt x\big)~,\quad$ and $\quad F(3)~=~\bigg[\dfrac2\pi~K\bigg(\dfrac{\sqrt{2-2~\sqrt{1-64x}}}2\bigg)\bigg]^2$. – Lucian Apr 11 '15 at 19:29
• Letting $~x=\bigg[\dfrac{\sin(2a)}{2^k}\bigg]^2,~$ with $~|a|<\dfrac\pi4,~$ the above expressions can be rewritten as $$F_0~=~\sec^2(2a),\quad F_1~=~\sec(2a),\quad F_2~=~\dfrac2\pi~K\Big(\sin(2a)\Big),\quad F_3~=~\bigg[\dfrac2\pi~K(\sin a)\bigg]^2.$$ – Lucian Jan 24 '17 at 5:48
• See another proof via hypergeometric functions at math.stackexchange.com/a/2152231/72031 – Paramanand Singh Feb 28 '17 at 4:33

We have: $$\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\,x^n = \frac{1}{\sqrt{1-x}},\tag{1}$$ $$\frac{1}{4^n}\binom{2n}{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left(\sin x\right)^{2n}\,dx \tag{2}$$ $$\sum_{n\geq 0}\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 x^n = \frac{2}{\pi}\,K(x)=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-x\sin^2\theta}}\tag{3}$$ hence by $(2)$ and $(3)$ it follows that: $$\begin{eqnarray*}\sum_{n\geq 0}\left(\frac{1}{4^n}\binom{2n}{n}\right)^3 &=& \frac{1}{\pi^2}\int_{-\pi}^{\pi}K(\sin^2 x)\,dx\\&=&\frac{4}{\pi^2}\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{1}{\sqrt{1-\sin^2\varphi\sin^2\theta}}\,d\theta\,d\varphi\\&=&\frac{4}{\pi^2}\,K\left(\frac{1}{2}\right)^2=\color{red}{\frac{\pi}{\Gamma\left(\frac{3}{4}\right)^4}},\tag{4}\end{eqnarray*}$$ so the original series equals $\displaystyle\color{purple}{-1+\frac{\pi}{\Gamma\left(\frac{3}{4}\right)^4}=0.39320392968567685918424626\ldots}.$
Footnote: this is just a very special case of the identity $(6)$ for the square of the complete elliptic integral of the first kind, plus the fact that $K(1/2)$ can be computed through the reflection and multiplication formulas for the $\Gamma$ function.
• @MathsFun: yes, the generalization is just: $$\sum_{n\geq 0}\left(\frac{1}{4^n}\binom{2n}{n}\right)^{2k+1}=\phantom{}_{2k+1} F_{2k}\left(\frac{1}{2},\ldots,\frac{1}{2};1,\ldots,1;1\right).$$ – Jack D'Aurizio Apr 11 '15 at 15:06
• @AdityaNarayanSharma: expand $K(\sin^2\theta)$ in terms of $\sin\theta$ through $(3)$. Integrate termwise through $(2)$: there is no Cauchy product. – Jack D'Aurizio Jan 14 '17 at 16:50