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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • 1
    $\begingroup$ 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
  • 3
    $\begingroup$ What an impressive looking solution!! Wallis would have been proud... :) $\endgroup$ – hypergeometric Apr 11 '15 at 15:04
  • 2
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.