A Ramanujan infinite series

Question

$$ 1-5\left(\frac{1}{2}\right)^3+9\left(\frac{1 \cdot 3}{2 \cdot 4}\right)^3-13\left(\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\right)^3+\cdots $$

I went on evaluating the above series and encountered that solving $\displaystyle \sum_{n\ge 0}\left(\binom{2n}{n}\right)^3x^n$ would suffice.

But how do we make a generating function for the third power of a central binomial coefficient using the fact $\displaystyle \sum_{n\ge 0}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$

The generating function is not algebraic. It can be expressed using a hypergeometric function: see OEIS sequence A002897. But that hypergeometric function is unlikely to help you. — Robert Israel, Jan 13 '17 at 05:23
The "power 2" generating function is in terms of elliptic integral and this one as you say in terms of hypergeometric functions, so there isn't any closed form I guess. But Ramanujan did find a value to the series I posted above. It was like $\frac{2}{\pi}$ as per my memory. — Aditya Narayan Sharma, Jan 13 '17 at 05:27

score 3 · Accepted Answer · answered Jan 13 '17 at 05:44

3

This is not an answer but just a result obtained using a CAS.

Let $$f_k=\sum_{n=0}^\infty \binom{2 n}{n}^kx^n$$ The following expressions have been obtained $$f_1=\frac{1}{\sqrt{1-4 x}}$$ $$f_2=\frac{2 }{\pi }K(16 x)$$ $$f_3=\frac{4 }{\pi ^2}K\left(\frac{1}{2} \left(1-\sqrt{1-64 x}\right)\right)^2$$ $$f_4=\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1;256 x\right)$$ $$f_5=\, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1,1;1 024 x\right)$$ where appear, for $k=2,3$, the complete elliptic integrals of the first kind and, for $k>3$, the generalized hypergeometric functions

answered Jan 13 '17 at 05:44

Claude Leibovici

260,315

"+1" , Thanks for the additional info , Is there any proof of these 5 over the web ? – Aditya Narayan Sharma Jan 13 '17 at 05:48
@AdityaNarayanSharma. I did not search for that. After your comment, just out of curiosity, I gave a CAs to eat ! Cheers. – Claude Leibovici Jan 13 '17 at 05:50
In the third case , the argument inside K is squared right ? And not the entire K-function ? – Aditya Narayan Sharma Jan 13 '17 at 05:53
@AdityaNarayanSharma. It is the entire function which is squared. – Claude Leibovici Jan 13 '17 at 08:10
@ClaudeLeibovici: Great minds think alike ! ;-) – Lucian Jan 13 '17 at 08:39
@Lucian. No merit at all ! Just a CAS made it. Cheers. – Claude Leibovici Jan 13 '17 at 08:42
3

Rewriting both my comment and your answer in terms of $~x=\bigg[\dfrac{\sin(2a)}{2^k}\bigg]^2,~$ with $~|a|<\dfrac\pi4,~$ we have $$F_0~=~\sec^2(2a),\quad F_1~=~\sec(2a),\quad F_2~=~\dfrac2\pi~K\Big(\sin(2a)\Big),\quad F_3~=~\bigg[\dfrac2\pi~K(\sin a)\bigg]^2.$$ Note: I've used the traditional modulus notation $K(k)$ for the argument of elliptic integrals, instead of the Mathematica one in terms the parameter $m=k^2.~$ – Lucian Jan 13 '17 at 09:33
@Lucian. This is very nice, for sure ! Thanks for providing it. Cheers. – Claude Leibovici Jan 13 '17 at 09:51
Simpler. Mathematica code: f[k_] := HypergeometricPFQ[Table[1/2, k], Table[1, k - 1], 4^k*x]; f[5] – Mariusz Iwaniuk Jan 17 '22 at 16:34

Paramanand Singh · Answer 2 · 2017-02-27T18:00:01.530

1

This is one of the simplest and famous series given by Ramanujan and it's value is $2/\pi$. Unfortunately Ramanujan's technique requires a reasonable amount of effort to understand. I have presented the proof for this series and it's friend $$\frac{4}{\pi}= 1+\frac{7}{4}\left(\frac{1}{2}\right)^{3}+\frac{13}{4^{2}}\left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}+\dots$$ in this post.

edited Feb 27 '17 at 18:00

answered Feb 27 '17 at 17:51

Paramanand Singh

87,309

+1 Thanks for the link, I knew one or two new identities – Aditya Narayan Sharma Feb 27 '17 at 18:42
@AdityaNarayanSharma: in case you want more there are some more series in the next blog post http://paramanands.blogspot.com/2012/03/modular-equations-and-approximations-to-pi-part-3.html – Paramanand Singh Feb 28 '17 at 04:15

rrogers · Answer 3 · 2022-01-17T14:51:28.573

Modifying @Claude Leibovici's answer a little.
$\frac{1}{\left(1-z\right)^{a}}=\,_{1}F_{0}\left(a;;z\right)={\displaystyle \sum_{n=0}^{\infty}\frac{a_{n}}{n!}z^{n}}$
and
$\frac{1}{\sqrt{1-4\cdot x}}={\displaystyle \sum_{n=0}^{\infty}\left(\begin{array}{c} 2n\\ n \end{array}\right)x^{n}}$
Let $z=4x,x=\frac{z}{4}$
$\frac{1}{\left(1-4x\right)^{a}}=\,_{1}F_{0}\left(a;;4x\right)={\displaystyle \sum_{n=0}^{\infty}\frac{a_{n}}{n!}\,\left(4x\right)^{n}}$
$4^{n}\frac{\left(\frac{1}{2}\right)_{n}}{n!}=\left(\begin{array}{c} 2n\\ n \end{array}\right)$
And we have the generating function for $\left(\begin{array}{c} 2n\\ n \end{array}\right)^{k}\left(4^{k}x\right)^{n}$
i.e.
$\,_{k}F_{k-1}\left({\displaystyle \prod_{i=1}^{k}\left(\frac{1}{2}\right)_n};{\displaystyle \prod_{i=1}^{k-1}\left(1\right)_n};4^{k}z\right)={\displaystyle \sum_{n=0}^{\infty}}\left(\begin{array}{c} 2n\\ n \end{array}\right)^{k}\left(4^{k}x\right)^{n}$
Where we “borrowed” the $n!=\left(1\right)_{n}$

A Ramanujan infinite series

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