Rapidly Converging Series for Particular Values of the Gamma Function

Question

C. H. Brown found some rapidly converging infinite series for particular values of the gamma function , $$\frac {\Gamma\left(\tfrac 13\right)^6}{12\pi^4}=\frac 1{\sqrt{10}}\sum\limits_{k=0}^{\infty}\frac {(6k)!}{k!^3(3k)!}\frac {(-1)^k}{(3\cdot160^3)^k}$$$$\frac {\Gamma\left(\tfrac 14\right)^4}{128\pi^3}=\frac 1{\sqrt u}\sum\limits_{k=0}^{\infty}\frac {(6k)!}{k!^3(3k)!}\frac {1}{\big(2\sqrt2\,u^3v^6\big)^k}$$

where $u=273+180\sqrt2,$ and $v=1+\sqrt2$.

Questions:

1. How did C. Brown derive these respective infinite formulas?
2. Can you derive similar formulas for different gamma functions?

I couldn't help but notice that these formulas share a similarity to the Chudnovsky Algorithm$$\frac 1\pi=12\sum\limits_{k=0}^{\infty}\frac {(6k)!}{k!^3(3k)!}\frac {545140134k+13591409}{640320^{k+1/2}}$$So perhaps Brown used the J-function and modular forms to derive the values?$$j(\tau)=\frac 1q+744+196884q+21493760q^2+\cdots$$

Answer 1

Back in 2011, I empirically found the general family for this type of formulas. The OP was right when he compared them to the Chudnovsky algorithm which uses the j-function $j(\tau)$.

---

I. j-function

The first has $j\big(\frac{1+\sqrt{-27}}{2}\big) = -3\times160^3.$

The second has $j(\sqrt{-16}) = 2\sqrt2\,(273+180\sqrt2)^3(1+\sqrt2)^6.$

A third should be $j(\sqrt{-6}) = 12^3\,(5+2\sqrt2)^3(1+\sqrt2)^2.$

---

II. Watson triple integrals

Brown's formulas can be connected to the "Watson triple integrals". (He missed the third one though.) Let,

\begin{align} I_1 &= \frac{1}{\pi^3} \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \frac{dx \, dy \, dz}{1-\cos x \cos y \cos z}\\ I_2 &= \frac{1}{\pi^3} \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \frac{dx \, dy \, dz}{3-\cos x \cos y -\cos x \cos z -\cos y \cos z}\\ I_3 &= \frac{1}{\pi^3} \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \frac{dx \, dy \, dz}{3-\cos x -\cos y -\cos z} \end{align}

then,

\begin{align} \frac{4^{4/3}}{3^2}\color{red}{I_2} &= \frac 1{\sqrt{10}}\sum\limits_{k=0}^{\infty}\frac {(6k)!}{k!^3(3k)!}\frac {(-1)^k}{(3\cdot160^3)^k} = \frac {\Gamma\left(\tfrac 13\right)^6}{12\pi^4}\\ \frac{1}{2^5}\color{red}{I_1} &= \frac 1{\sqrt u}\sum\limits_{k=0}^{\infty}\frac {(6k)!}{k!^3(3k)!}\frac {1}{\big(2\sqrt2\,u^3\,(1+\sqrt2)^6\big)^k} = \frac {\Gamma\left(\tfrac 14\right)^4}{128\pi^3}\\ \frac{1}{2^{1/2}}I_3 &= \frac 1{\sqrt w}\sum\limits_{k=0}^{\infty}\frac {(6k)!}{k!^3(3k)!}\frac {1}{\big(12^3w^3\,(1+\sqrt2)^2\big)^k} = \frac {\Gamma\left(\tfrac 1{24}\right)\Gamma\left(\tfrac 5{24}\right)\Gamma\left(\tfrac 7{24}\right)\Gamma\left(\tfrac {11}{24}\right)}{32\sqrt3\,\pi^3} \end{align}

where $u=273+180\sqrt2,$ and $w=5+2\sqrt2$.

---

III. Gamma functions

For the first, given $\tau = \big(\frac{1+\sqrt{-d}}{2}\big)$ where $d>3$, a general formula is the simple sum-product identity,

$$\frac1{\big({-j(\tau)}\big)^{1/6}}\sum_{n=0}^\infty \frac{(6n)!}{(3n)!(n!^3)} \frac1{\big(j(\tau)\big)^n} = \frac1{(2\pi)d}\prod_{m=1}^{d-1} \Big[\Gamma\big(\tfrac m d\big)\Big]^{\big(\frac{-d}{m}\big)}$$

where the gamma functions's exponent $\big(\frac{-d}{m}\big)$ is the Kronecker symbol. For example, let $d = 7$, so $j\big(\frac{1+\sqrt{-7}}{2}\big)=-15^3$. Thus,

$$\frac1{\big({15^3}\big)^{1/6}}\sum_{n=0}^\infty \frac{(6n)!}{(3n)!(n!^3)} \frac1{\big({-15^3}\big)^n} = \frac1{7(2\pi)}\frac{\Gamma\big(\tfrac17\big)\Gamma\big(\tfrac27\big)\Gamma\big(\tfrac47\big)}{\Gamma\big(\tfrac37\big)\Gamma\big(\tfrac57\big)\Gamma\big(\tfrac67\big)} $$

and so on.

Answer 2

That Wikipedia link points to http://www.iamned.com/math/ which has a large number of amazing formulae.

That, in turn, points to http://iamned.com/math/infiniteseries.pdf titled "An Algorithm for the Derivation of Rapidly Converging Infinite Series for Universal Mathematical Constants".

Answer 3

We have derived some of this values you are searching in this question: Method to obtain values of Gauss hypergeometric function.

That allows to get: $$\sum_{n=0}^{\infty}\frac{(3n)!(2n)!(-1)^n}{n!^5}\left(\frac{1}{8640} \right)^n=\frac{9\cdot 5^{2/3}\cdot\Gamma(1/3)^6}{100\pi^4},\tag{1}$$ $$\sum_{n=0}^{\infty}\frac{(4n)!(-1)^n}{n!^4}\left(\frac{1}{6635520} \right)^n=\frac{3\cdot 5^{1/4}\cdot\Gamma(1/4)^4}{25 \pi^3}.\tag{2}$$ Using Clausen's formula: $$\left(\sum_{n=0}^{\infty}\frac{f(a)f(b)}{f(2b)n!}x^n\right)^2=(1-x)^{-a}\sum_{n=0}^{\infty}\frac{f(a)f(b)f(2b-a)}{f(b+1/2)f(2b)n!}\left(\frac{x^2}{4(x-1)} \right)^n\tag{3}$$ Where $f(x)=\frac{\Gamma{(x+n)}}{\Gamma{(x)}}=(x)_{n},$ is the Pochhammer symbol. Some values are obtained from Ebisu's work. There are formulas connecting the four known theories in which binomial coefficients are of the form: $$\frac{(2n)!^3}{n!^6}\hspace{.5cm}\frac{(3n)!(2n)!}{n!^5}\hspace{.5cm}\frac{(4n)!}{n!^4}\hspace{.5cm}\frac{(6n)!}{(3n)!n!^3} \tag{4}$$ Perhaps one of the most simple is: $$\sum_{n=0}^{\infty}\frac{(-1)^n(2n)!^3}{2^{9n}n!^6}=\frac{\Gamma{(1/4)}^4}{4\sqrt{2}\pi^3}\tag{5}$$ which you can find here: Integrals of elliptic integrals. We also arrived to: $$\sum_{n=0}^{\infty}\frac{(-1)^n(2n)!^3(2n+1)}{2^{9n}n!^6}=\frac{2\sqrt{2}}{3\pi}+\frac{\Gamma(\frac{1}{4})^4}{6\sqrt{2}\pi^3}\tag{6}$$ and $$\sum_{n=0}^{\infty}\frac{(-1)^n(6n+1)(2n)!^3}{2^{9n}n!^6}=\frac{2\sqrt{2}}{\pi}.\tag{7}$$ Another example is obtained here also here On Ramanujan's fastest series. in an elementary way: $$\sum_{n=0}^{\infty}\frac{(4n)!}{648^n n!^4}=\frac{3\Gamma{(\frac{1}{4})^4}}{16\pi^3}\tag{8}$$ and the associated series for $1/\pi$: $$\sum_{n=0}^{\infty}\frac{(7n+1)(4n)!}{648^n n!^4}=\frac{9}{2\pi}.\tag{9}$$

A very interesting paper by Zudilin https://arxiv.org/abs/1210.0269 (Lost in translation) gives the links between the four theories. So $(5)$ and $(8)$ and $(7)$ and $(9)$ are the same in the sense of Zudilin's paper.

Finally Piezas's answer is in fact Chowla-Selberg formula with gives the explicit evaluation of the complete elliptic of first kind in terms of pi and gamma functions.