Closed form for $\sum_{n=0}^\infty\frac{\Gamma\left(n+\tfrac14\right)}{2^n\,(4n+1)^2\,n!}$ I was experimenting with hypergeometric-like series and discovered the following conjecture (so far confirmed by more than $5000$ decimal digits):
$$\sum_{n=0}^\infty\frac{\Gamma\!\left(n+\tfrac14\right)}{2^n\,(4n+1)^2\,n!}\stackrel{\color{gray}?}=$$
$$\frac{\Gamma\!\left(\tfrac14\right)\sqrt[4]2}{192}\left[\vphantom{\huge|}6\sqrt{2}\left(2\pi\ln2-\ln^22-8\operatorname{Li}_2\left(\tfrac1{\sqrt2}\right)\right)+3\psi^{(1)}\!\left(\tfrac18\right)-48G+\left(\vphantom{\large|}7\sqrt2-6\right)\pi^2\right]$$
where $G$ is the Catalan constant, $\operatorname{Li}_2(x)$ is the dilogarithm and $\psi^{(1)}(x)$ is the trigamma function. Could you suggest any ideas how to prove it?

To see what approach I use to find conjectures like this, see my another question.

Update: I've found a generalization of this conjecture. See the corresponding Mathematica expression here. Hopefully, it can be simplified.
 A: (Too long for a comment.) Note that,
$$\sum_{n=0}^\infty\frac{\Gamma\!\left(n+\tfrac14\right)}{2^n\,(4n+1)^2\,n!}=A=B$$
$$A=\frac{\Gamma\!\left(\tfrac14\right)\sqrt[4]2}{192}\left[\vphantom{\huge|}6\sqrt{2}\left(2\pi\ln2-\ln^22-8\operatorname{Li}_2\left(\tfrac1{\sqrt2}\right)\right)+3\psi^{(1)}\!\left(\tfrac18\right)-48G+\left(\vphantom{\large|}7\sqrt2-6\right)\pi^2\right]$$
$$B=\frac{\Gamma\!\left(\tfrac14\right)\sqrt[4]2}{192}\left[\vphantom{\huge|}6\sqrt{2}\left(2\pi\ln2-\ln^22-8\operatorname{Li}_2\left(\tfrac1{\sqrt2}\right)\right)\color{red}-3\psi^{(1)}\!\left(\color{red}{\tfrac58}\right)\color{red}+48G+\left(\vphantom{\large|}7\sqrt2\color{red}+6\right)\pi^2\right]$$
Since,
$$\psi^{(1)}\!\left(\tfrac18\right)+\psi^{(1)}\!\left(\tfrac78\right)=2\pi^2(2+\sqrt2)$$
$$\psi^{(1)}\!\left(\tfrac38\right)+\psi^{(1)}\!\left(\tfrac58\right)=2\pi^2(2-\sqrt2)$$
Or in general,
$$\psi^{(1)}\!\left(k\right)+\psi^{(1)}\!\left(1-k\right)=\pi^2\csc^2(k\pi)$$
then one can use any of the arguments $\tfrac18,\tfrac38,\tfrac58,\tfrac78$.
