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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.

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

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    $\begingroup$ So the question is really hard and solution is tricky. Just one question @Jack D'Aurizio here the power term could be generalized (using 5 instead of 3) and solved in same way. $\endgroup$ – Maths Fun Apr 11 '15 at 13:55
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    $\begingroup$ What an impressive looking solution!! Wallis would have been proud... :) $\endgroup$ – hypergeometric Apr 11 '15 at 15:04
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    $\begingroup$ @MathsFun: yes, the generalization is just: $$\sum_{n\geq 0}\left(\frac{1}{4^n}\binom{2n}{n}\right)^{2k+1}=\phantom{}_{2k+1} F_{2k}\left(\frac{1}{2},\ldots,\frac{1}{2};1,\ldots,1;1\right).$$ $\endgroup$ – Jack D'Aurizio Apr 11 '15 at 15:06
  • $\begingroup$ @JackD'Aurizio I just want to know how you combined (2) & (3) to get the main series. I mean how to multiply a expression with a power series ? It isn't the cauchy product $\endgroup$ – Aditya Narayan Sharma Jan 14 '17 at 8:15
  • $\begingroup$ @AdityaNarayanSharma: expand $K(\sin^2\theta)$ in terms of $\sin\theta$ through $(3)$. Integrate termwise through $(2)$: there is no Cauchy product. $\endgroup$ – Jack D'Aurizio Jan 14 '17 at 16:50

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