A conjectural closed form for $\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}$

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.

partial answer

Use the method of LINK

Since $$ \int_0^1 t^n (1-t)^{2n}\;dt = \frac{n!(2n)!}{(3n+1)!} , $$ If we write $$ S(x) = \sum_{n=0}^\infty \frac{n!(2n)!}{(3n+2)!}\;x^{3n+2} $$ then $$ S(x) = \int_0^1 \left(\sum_{n=0}^\infty t^n(1-t)^{2n} \frac{x^{3n+2}}{3n+2}\right)dt $$ and $S = S(1)$.
But the derivative with respect to $x$ of $\sum_{n=0}^\infty t^n(1-t)^{2n} \frac{x^{3n+2}}{3n+2}$ is a geometric series. Its sum is a rational function, which may be integrated (with some work or a CAS). Plug in $x=1$. Result (if I copied right):

$$ S = \int_0^1\frac{F(t)}{18(1-t)^{4/3}t^{2/3}}\;dt $$

where

$$ F(t) = \pi \,\sqrt {3}-6\,\sqrt {3}\arctan \left( 2/\sqrt {3}\cdot \left( 1-t \right) ^{2/3}{t}^{1/3}+1/ \sqrt {3} \right) \\ -6\,\ln \left( ({1-t})^{1/3} {t}^{2/3} -\left( 1-t \right) t \right) +3\,\ln \left( t \left( 1-t \right) ^{2}+ \left( 1-t \right) ^{4/3}{t}^{2/3}+ \left( 1-t \right) ^{2/3}{t}^{1/3} \right) \\ +6\,\ln \left( ({1-t})^{1/3}{t}^{2/3} \right) -3\,\ln \left( \left( 1-t \right) ^{2/3}{t}^{1/3} \right) $$

Alternatively, one can find a simple and elementary solution in logarithms as, $$\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}= -x_1\ln(1-x_1)-x_2\ln(1-x_2)-x_3\ln(1-x_3) = 0.5179778\dots$$ where the $x_n$ are the three roots of $x^3-x+1=0$ and which is the minpoly of the negated plastic constant. See also this post.