Is there a simpler closed form for $\sum_{n=1}^\infty\frac{(2n-1)!!\ (2n+1)!!}{4^n\ (n+2)\ (n+2)!^2}$ I have the following infinite sum that can be expressed in terms of the generalized hypergeometric function:
$$\sum_{n=1}^\infty\frac{(2n-1)!!\ (2n+1)!!}{4^n\ (n+2)\ (n+2)!^2}=\frac{31}8-4\times{_4F_3}\left(-\frac12,\frac12,1,1;\ 2,2,2;\ 1\right)\\\ \\\approx0.008749644047541935203478962326551903908774780849356243615274...$$
I wonder if it can be expressed in terms of simpler functions and well-known mathematical constants.
 A: Ookay... Consider just the hypergeometric function and its closed form.
Take 16.5.2 from DLMF (with $a_0=1/2$, $b_0=2$) and write (I used Mathematica to substitute the special form for the hypergeometric function in the integrand; I don't really know how to do it by hand):
$$ F(-1/2,1/2,1,1;2,2,2;1) = \int_0^1 \frac{8 \sqrt{1-t} \left(4-4 \sqrt{1-t}+t\sqrt{1-t}-\log8+3 \log(1+\sqrt{1-t})\right)}{9 \pi  t^{3/2}}\,dt. $$
Mathematica can then do the integral in closed form to give
$$\frac{1}{9 \pi }8 \left(\frac{70}{3}+\frac{11 i \pi ^2}{4}+\pi  \left(-7+\log 512-6\log\left(1+\frac{1+i}{\sqrt{2}}\right)\right)+24 i \mathrm{Li}_2\left(-\frac{1+i}{\sqrt{2}}\right)-24 i \mathrm{Li}_2\left(1-\frac{1+i}{\sqrt{2}}\right)\right),$$
where $\mathrm{Li}_2$ is the polylogarithm.
Of this we take the real part only and do some more FunctionExpand:
$$-\frac{56}{9}+\frac{560}{27 \pi }-\frac{16 C}{3 \pi }+\frac{64 \Im\left(\mathrm{Li}_2(1-(-1)^{1/4})\right)}{3 \pi }+\frac{8 \log512}{9}-\frac{8}{3} \log\left(\frac{1}{2}+\left(1+\frac{1}{\sqrt{2}}\right)^2\right)+\frac{\psi_1\left(\frac{1}{8}\right)}{3 \sqrt{2} \pi }+\frac{\psi_1\left(\frac{3}{8}\right)}{3 \sqrt{2} \pi }-\frac{\psi_1\left(\frac{5}{8}\right)}{3 \sqrt{2} \pi }-\frac{\psi_1\left(\frac{7}{8}\right)}{3 \sqrt{2} \pi }, $$
where $\psi_1$ is the polygamma function and $C$ is the Catalan constant.
Now, the polylogarithm term there can be simplified using the identity (DLMF 25.12.6)
$$ \mathrm{Li}_2(x)+\mathrm{Li}_2(1-x)=\frac{\pi^2}{6}-\log x\log(1-x), $$
because $\mathrm{Li}_2((-1)^{1/4})$ is simpler.
After a further FunctionExpand (which gets rid of the polygamma functions also), ComplexExpand to get the real part, and FullSimplify to simplify the expression, the answer is
$$ -\frac{56}{9}+\frac{560}{27 \pi }-\frac{32C}{3 \pi }+\frac{16}{3} \log2$$
A: I can give a partial answer:
My program based on the TranscendentalRecognize algorithm using a wide set of commonly occuring constants after several hours of work discovered the following inequality:
$$\Bigg|\frac{16\ln2}{3}-\frac{32C}{3\pi}+\frac{560}{27\pi}-\frac{56}{9}-{_4F_3}\left(-\frac{1}{2},\frac{1}{2},1,1;2,2,2;1\right)\Bigg|<10^{-1000},$$
where $C$ is the Catalan constant.
I have no idea if the actual difference is exact $0$, or how to (dis-)prove it.
