Let

S=\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!},\tag1

its numeric value is approximately S \approx 0.517977853388534047…{}^{[more\ digits]}S can be represented in terms of the generalized hypergeometric function:

S={_3F_2}\left(\frac12,1,1;\ \frac43,\frac53;\ \frac4{27}\right)\cdot\frac12.\tag2

Let \sigma be the closed-form expression constructed from integers and elementary functions as follows:

\sigma=3\,\alpha\,\ln(2\,\alpha+1)-\sqrt{\beta\,}\arccos\gamma,\tag3

where

\alpha=\frac{\sqrt[3]{3\,}}{6\,\sqrt[3]{2\,}}\left(\sqrt[3]{9-\sqrt{69}}+\sqrt[3]{9+\sqrt{69}}\right),\tag4

\beta=\frac1{4\,\sqrt[3]{2\,}}\left(\sqrt[3]{25+3\,\sqrt{69}}+\sqrt[3]{25-3\,\sqrt{69}}\right)-\frac12,\tag5

\gamma=\frac1{6\,\sqrt[3]{2\,}}\left(\sqrt[3]{57\,\sqrt{69}-459}-\sqrt[3]{57\,\sqrt{69}+459}\right)+\frac12\tag6

are the unique real roots of the following cubic equations:

8\,\alpha^3-2\,\alpha-1=0,\tag7

64\,\beta^3+96\,\beta^2+36\,\beta-23=0,\tag8

8\,\gamma^3-12\,\gamma^2+16\,\gamma+11=0.\tag9

Equivalently,

\sigma = \frac{3\,p}{2}\,\ln\big(p+1\big)-\frac{1}{2}\sqrt{\frac{3-p}{p}}\arccos\Big( \frac{p-6}{6p+2}\Big)\tag{10}

where p is theplastic constantor the real root of

p^3-p-1=0\tag{11}

It can be numerically checked that the following inequality holds:

\Big|S-\sigma\Big|<10^{-10^5},\tag{12}

I conjecture that the actual difference is the exact zero, and thus S has an elementary closed form:

\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}\stackrel?=3\,\alpha\,\ln(2\,\alpha+1)-\sqrt{\beta\,}\arccos\gamma,\tag{13}

I am asking for you help in proving this conjecture.

**Answer**

We show that the sum equals

\int_0^1 \frac{2-3x}{1-x^2+x^3} dx.

This integral is "elementary", but requires expanding the integrand

in partial fractions, which in turn requires all the solutions of

the cubic polynomial in the denominator; so if one insists on

writing everything in radicals then the answer is bound to be complicated.

The "conjecture" is surely correct (10^5 digits is more than enough

for moral certainty, especially since \alpha,\beta,\gamma are all

in the field generated by the real root of 1-x^2+x^3), though

it may be an unpleasant and unrewarding exercise to check that

the partial-fraction integration yields an equivalent answer.

(One also wonders how one could possibly "conjecture" such an answer

without some sense of where to look...)

The key is to write each term n! (2n)! / (3n+2)! in terms of the

beta integral

a!b!/(a+b+1)! = B(a+1,b+1) = \int_0^1 x^a (1-x)^b dx. Here we write

n! (2n)! / (3n+2)! = B(2n+1, n+2) / (n+1), and sum over n to get

\sum_{n=0}^\infty \frac{n! (2n)!} {(3n+2)!}

= \int_0^1 \sum_{n=0}^\infty \frac{(x^2-x^3)^{n+1}}{n+1} \frac{dx}{x^2}

= -\int_0^1 \log(1-x^2+x^3) \frac{dx}{x^2}.

(We easily justify the interchange of infinite sum and definite integral

because all integrands are positive on 0<x<1.)

We can now integrate by parts to remove the logarithm:

-\int_0^1 \log(1-x^2+x^3) \frac{dx}{x^2}

= \int_0^1 \log(1-x^2+x^3) \phantom. d\left(\frac{1}{x}\right)

= \int_0^1 \frac1x d(\log(1-x^2+x^3)),

in which the integrand simplifies to (2-3x)/(1-x^2+x^3),

**QED**.

**Attribution***Source : Link , Question Author : Vladimir Reshetnikov , Answer Author : Noam D. Elkies*