How can I prove the following conjectured identity? $$\mathcal{S}=\sum_{n=1}^\infty\frac{(4\,n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^n}\stackrel?=\frac{\sqrt3}{2\,\pi}\left(2\sqrt{\frac8{\sqrt\alpha}-\alpha}-2\sqrt\alpha-3\right),$$ where $$\alpha=2\sqrt[3]{1+\sqrt2}-\frac2{\sqrt[3]{1+\sqrt2}}.$$ The conjecture is equivalent to saying that $\pi\,\mathcal{S}$ is the root of the polynomial $$256 x^8-6912 x^6-814752 x^4-13364784 x^2+531441,$$ belonging to the interval $-1<x<0$.

The summand came as a solution to the recurrence relation $$\begin{cases}a(1)=-\frac{81\sqrt3}{512\,\pi}\\\\a(n+1)=-\frac{9\,(2n+1)(4n+1)(4 n+3)}{32\,(n+1)(3n+2)(3n+4)}a(n)\end{cases}.$$ The conjectured closed form was found using computer based on results of numerical summation. The approximate numeric result is $\mathcal{S}=-0.06339748327393640606333225108136874...$ (click to see 1000 digits).

  • 2
    $\begingroup$ Wow! Either there's some slick trick there, or some hard development of something, or else...a miracle's needed here! Where does this come from, context, what have you done so far...? $\endgroup$
    – DonAntonio
    May 25, 2013 at 22:51
  • $\begingroup$ Where does this bizarre formula(s) come from? I'd see what happens when the gamma values are expressed in terms of factorial powers (by using the recurrence), it might end up looking like a multinomial coefficient of sorts... $\endgroup$
    – vonbrand
    May 25, 2013 at 22:55
  • $\begingroup$ I think expressing $(4n)!=\Gamma(4n+1)$, and using 4-multiplication formula for the gamma function, this becomes some $_pF_q$-function evaluated at particular values of parameters and independent variable (and there one has a lot of funny formulas to play with). It doesn't seem to me to be a real question, like many others of the same kind recently. However, I will not develop this further - the last time my answer ended with a warning from moderator team. $\endgroup$ May 25, 2013 at 23:02
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    $\begingroup$ Indeed. It's something Ramanujan would come up with. Wolframalpha says the ratio test is inconclusive, but can give a numerical value. $\endgroup$ May 25, 2013 at 23:04
  • $\begingroup$ @O.L. I'm inclined to agree about this question, and I would also really like to know how, given a number, one finds precisely the integer polynomial of 8-th degree that it turns out to be the root of. There must be so many polynomials of not too large a degree with approximately the right roots, especially given the magnitude of coefficients. $\endgroup$
    – Kirill
    May 26, 2013 at 2:50

2 Answers 2


According to Mathematica, the sum is $$ \frac{3}{\Gamma(\frac13)\Gamma(\frac23)}\left( -1 + {}_3F_2\left(\frac14,\frac12,\frac34; \frac23,\frac43; -1\right) \right). $$

This form is actually quite straightforward if you write out $(4n)!$ as $$ 4^{4n}n!(1/4)_n (1/2)_n (3/4)_n $$ using rising powers ("Pochhammer symbols") and then use the definition of a hypergeometric function.

The hypergeometric function there can be handled with equation 25 here: http://mathworld.wolfram.com/HypergeometricFunction.html: $$ {}_3F_2\left(\frac14,\frac12,\frac34; \frac23,\frac43; y\right)=\frac{1}{1-x^k},$$ where $k=3$, $0\leq x\leq (1+k)^{-1/k}$ and $$ y = \left(\frac{x(1-x^k)}{f_k}\right)^k, \qquad f_k = \frac{k}{(1+k)^{(1+1/k)}}. $$

Now setting $y=-1$, we get the polynomial equation in $x$ $$ \frac{256}{27} x^3 \left(1-x^3\right)^3 = -1,$$ which has two real roots, neither of them in the necessary interval $[0,(1+k)^{-1/k}=4^{-1/3}]$, since one is $-0.43\ldots$ and the other $1.124\ldots$. However, one of those roots, $x_1=-0.436250\ldots$ just happens to give the (numerically at least) right answer, so never mind that.

Also, note that $$ \Gamma(1/3)\Gamma(2/3) = \frac{2\pi}{\sqrt{3}}. $$

The polynomial equation above is in terms of $x^3$, so we can simplify that too a little, so the answer is that the sum equals $$ \frac{3^{3/2}}{2\pi} \left(-1+(1-z_1)^{-1}\right), $$ where $z_1$ is a root of the polynomial equation $$ 256z(1-z)^3+27=0, \qquad z_1=-0.0830249175076244\ldots $$ (The other real root is $\approx 1.42$.)

How did you find the conjectured closed form?

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    $\begingroup$ I calculated the sum in Mathematica and then tried RootApproximant[N[(3 Sqrt[3] (-1 + HypergeometricPFQ[{1/4, 1/2, 3/4}, {2/3, 4/3}, -1]))/(2 Pi), 100]] - it returned a polynomial of degree 33, which failed to match the expression numerically when I used a higher precision. Then I guessed it could work if the factor 1\Pi were excluded and tried RootApproximant[N[(3 Sqrt[3] (-1 + HypergeometricPFQ[{1/4, 1/2, 3/4}, {2/3, 4/3}, -1]))/2, 100]] - it returned a polynomial of degree 8 (the one I mentioned in the question), which agreed with the expression numerically to a very high precision. $\endgroup$
    – Hanna K.
    May 26, 2013 at 3:30
  • $\begingroup$ RootApproximant is really powerful tool to find algebraic closed forms if you have enough precision. Sometimes it works better if you explicitly set the maximum degree too a reasonable value (4, 6 or 8) as the second argument. $\endgroup$
    – Hanna K.
    May 26, 2013 at 3:38
  • $\begingroup$ Thanks for explaining, that's quite interesting. $\endgroup$
    – Kirill
    May 26, 2013 at 4:11

This is a very interesting question. Since it is different from similar question of this kind in that a new technique is involved I will provide the answer. We have: \begin{eqnarray} &&\sum\limits_{n=1}^\infty \frac{(4n)!}{\Gamma(n+\frac{2}{3}) \Gamma(n+\frac{4}{3})(n!)^2 (-256)^n}=\\ && \frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!} \sum\limits_{n=1}^\infty \binom{4 n}{n} \cdot \underbrace{\frac{1}{n+\frac{1}{3}}}_{\int\limits_0^1\theta^{n-\frac{2}{3}} d\theta}\cdot \left( -\frac{3^3}{4^4}\right)^n=\\ &&\frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!}\int\limits_0^1 \theta^{-\frac{2}{3}} \cdot \left\{ \frac{x[\theta](1+\theta \frac{27}{256}x[\theta]^3)}{1+4\theta \frac{27}{256}x[\theta]^3}-1\right\} d\theta=\\ &&\frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!}\left\{-\frac{4 \cdot 2^{\frac{2}{3}}}{3}\cdot \int\limits_1^{x[1]} \frac{1}{x^{\frac{4}{3}}} \cdot \frac{1}{(1-x)^{\frac{2}{3}}}dx-3\right\}\\ &&\frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!}\left\{4 \cdot 2^{\frac{2}{3}}(\frac{1-x[1]}{x[1]})^{\frac{1}{3}}-3\right\}\\ &&\frac{\sqrt{3}}{2 \pi}\left\{4 \cdot 2^{\frac{2}{3}}(\frac{1-x[1]}{x[1]})^{\frac{1}{3}}-3\right\}\\ \end{eqnarray} In the first line we used elementary properties of factorials. In the second line we used my answer to Closed form solutions for a family of hypergeometric sums. . Here the function$x[\theta]$ is defined as a solution to the following polynomial equation: \begin{equation} 1-x[\theta] - \theta \frac{27}{256} x[\theta]^4=0 \end{equation} where out of the four different solutions we take the solution that is the closest to unity. In the third line we substituted for $x=x[\theta]$, in the fourth line we evaluated the integral and finally in the fifthe line we simplified the result. Here : \begin{equation} x[1]=\frac{2 \sqrt{2+\left(1+\sqrt{2}\right)^{2/3} \left(4 \sqrt{\frac{2}{1+\sqrt{2}-\sqrt[3]{1+\sqrt{2}}}}-2\right)}-2 \sqrt{2 \left(\left(1+\sqrt{2}\right)^{2/3}-1\right)}}{3 \sqrt[6]{1+\sqrt{2}}} \end{equation}


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