TITLE: Is there a simpler closed form for $\sum_{n=1}^\infty\frac{(2n-1)!!\ (2n+1)!!}{4^n\ (n+2)\ (n+2)!^2}$
QUESTION [24 upvotes]: 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.

REPLY [3 votes]: Those calculations are really addictive.. Now, this sum is doable because the term in the sum can be written as a product of a binomial factor, a beta function and a rational function . Let me show how it goes. Firstly we have:
\begin{eqnarray}
(2n-1)!!&=& 2^n \frac{(n-\frac{1}{2})!}{(-\frac{1}{2})!}\\
(2n+1)!!&=& 2^n \frac{(n+\frac{1}{2})!}{(+\frac{1}{2})!}
\end{eqnarray}
Therefore the sum in question reads:
\begin{eqnarray}
&&\sum\limits_{n=1}^\infty \frac{(2n-1)!! (2n+1)!!}{4^n (n+2)[(n+2)!]^2}=\\
&&\sum\limits_{n=1}^\infty \frac{(n-\frac{1}{2})!}{(-\frac{1}{2})!} \frac{(n+\frac{1}{2})!}{(+\frac{1}{2})!} \cdot \frac{1}{(n+2)[(n+2)!]^2}\\
&&=\sum\limits_{n=1}^\infty \binom{n-\frac{1}{2}}{n} \cdot \frac{(n+\frac{1}{2})! (+\frac{1}{2})!}{[(+\frac{1}{2})!]^2(n+2)!} \cdot \frac{1}{(n+2)^2 (n+1)}\\
&&= \sum\limits_{n=1}^\infty (-1)^n \binom{-\frac{1}{2}}{n} \cdot \frac{1}{[(+\frac{1}{2})!]^2} \cdot \int\limits_0^1 t^{n+\frac{1}{2}} (1-t)^{\frac{1}{2}} dt \cdot \frac{1}{(n+2)^2 (n+1)}\\
&&= \frac{16}{3 \pi} \sum\limits_{n=1}^\infty (-1)^n\binom{\frac{3}{2}}{n+2} \cdot \int\limits_0^1 t^{n+\frac{1}{2}} (1-t)^{\frac{1}{2}} dt \cdot \underbrace{\frac{1}{(n+2)}}_{\int\limits_0^1 \theta^{n+1} d\theta}\\
&&=\frac{16}{3 \pi} \int \limits_0^1 t^{\frac{1}{2}} (1-t)^{\frac{1}{2}} \cdot \int\limits_0^1 \theta\underbrace{\left\{ \sum\limits_{n=1}^\infty \binom{\frac{3}{2}}{n+2} (-\theta t)^n\right\}}_{\frac{-3 t^2 \theta ^2+12 t \theta -8 t \theta  \sqrt{1-t \theta }+8 \sqrt{1-t \theta }-8}{8 t^2 \theta ^2}} d\theta dt\\
&&= \int\limits_0^1 (1-t)^{\frac{1}{2}} \cdot \\
&&\left( \frac{\left(-9 t-32 \sqrt{1-t}+72\right) t+32 \left(4 \sqrt{1-t}-4+\log (8)\right)-48 \log (t)-96 \tanh ^{-1}\left(\sqrt{1-t}\right)}{9 \pi  t^{3/2}}\right) dt =\\
&&\frac{31}{8}-\frac{128}{27 \pi} + \int\limits_0^1 (1-t)^{\frac{1}{2}}\cdot\left(\frac{32 \left(4 \sqrt{1-t}-4+\log (8)\right)-48 \log (t)-96 \tanh ^{-1}\left(\sqrt{1-t}\right)}{9 \pi  t^{3/2}} \right)dt=\\
&&\frac{1303}{72}-\frac{1664}{27 \pi} + \int\limits_0^1 (1-t)^{\frac{1}{2}}\cdot \left( \frac{-48 \log (t)-96 \tanh ^{-1}\left(\sqrt{1-t}\right)+32 \log (8)}{9 \pi  t^{3/2}}\right) dt=\\
&&\frac{1303}{72}-\frac{1664}{27 \pi} + \frac{32 (4 C-2+\pi -\pi  \log (4))}{3 \pi }=\\
&&\frac{128 C}{3 \pi }+\frac{2071}{72}-\frac{2240}{27 \pi }-\frac{32 \log (4)}{3}
\end{eqnarray}
which matches the answer given by Kirill.