Proving that $\sum_{k=0}^\infty\frac{2^{-5k}(6k+1)((2k-1)!!)^3}{4(k!)^3} = {1\over\pi}$

Question

While trying to prove that $$(1)\qquad x\sum_{k=0}^\infty\frac{2^{-5k}(6k+1)((2k-1)!!)^3}{4(k!)^3} = 1 \implies x=\pi$$

I got to a point, using W|A, where I have to prove that $$\color{red}{(2)\qquad \pi = -\frac{8 \left (\sqrt{16-3 K\left ({1\over4} (2-\sqrt{3})\right )^2\, _3 F_2\left ({3\over2}, {3\over2}, {3\over2};\,2, 2;{1\over4}\right )}-4\right )}{\left (3\, _3 F_2\left ({3\over2}, {3\over2}, {3\over2};\,2, 2;\,{1\over4}\right )\right )}}$$

Where $K(x)$ is the complete elliptic K function, and $_xF_y(a_1 ... a_p;b_1 ... b_q; z)$ is the generalized hypergeometric function.

IMPORTANT: As user153012 pointed out, there is likely an error in $(2)$, as I translated it from mathematica to Latex by hand, so I'll just post the mathematica query so that there is no misunderstanding:

-8 (-4 + Sqrt[16 - 3 EllipticK[(2 - Sqrt[3])/4]^2 HypergeometricPFQ[{3/2, 3/2, 3/2}, {2, 2}, 1/4]]))/(3 HypergeometricPFQ[{3/2, 3/2, 3/2}, {2, 2}, 1/4])

Here's a LINK to the query (If W|A show a message similar to "Wolfram|Alpha doesn't know how to interpret your input", just recompute the query by either refreshig the page or pressing the orange and white $=$ sign).

I will be correcting this error asap.

Is there a way to prove $(2)$ to be true, or another way to prove $(1)$ to be true (by eliminating the double factorial? I haven't had any success in doing this myself.)?

Thanks.

For me the RHS of $(2)$ is not $\pi$. It is $3.0135849404\dots$ — user153012, Oct 29 '14 at 17:43
@user153012 I've addressed the issue as best I can at the moment. Thanks for noticing. — JohnWO, Oct 29 '14 at 19:16
Like in the movie "The Thing", only that the thing here is that summation. — Pedro, Oct 29 '14 at 21:47
@PedroTamaroff (+1 for Carpenter reference)It got nothing on Ramanujan’s infinite series of ${1\over\pi}$... :P — JohnWO, Oct 30 '14 at 00:23
Hint: $~(2k-1)!!\cdot(2k)!!=(2k)!~$ and $~(2k)!!=2^k~k!~$ Then use a trick similar to this. — Lucian, Oct 30 '14 at 16:37

score 4 · Accepted Answer · edited Apr 13 '17 at 12:21

The formula,

$$\sum_{k=0}^\infty\frac{2^{-5k}(6k+1)((2k-1)!!)^3}{4(k!)^3} = \frac{1}{\pi}\tag1$$

or its equivalent form,

$$\sum_{k=0}^\infty \frac{(2k)!^3}{k!^6}\frac{6k+1}{(2^8)^{k+1/2}} = \frac{4}{\pi}\tag2$$

and the similar,

$$\sum_{k=0}^\infty (-1)^k \frac{(2k)!^3}{k!^6}\frac{6k+1}{(2^9)^{k+1/2}} = \frac{8}{\pi}\tag3$$

belong to the Ramanujan-type family,

$$\sum_{k=0}^\infty \frac{(2k)!^3}{k!^6}\frac{Ak+B}{(C)^{k+1/2}} = \frac{1}{\pi}\tag4$$

where $A,B,C$ are algebraic numbers. There are only four formulas such that $A,B,C$ are integers. The complete list can be found here.

The general case has already been proven by either the Borweins, Berndt, or Guillera. For the specific details, their papers can be found online. Raymond Manzoni has given a list in this post.

Proving that $\sum_{k=0}^\infty\frac{2^{-5k}(6k+1)((2k-1)!!)^3}{4(k!)^3} = {1\over\pi}$

1 Answers1