How to evaluate
$$\sum_{n=0}^{\infty }\sum_{k=1}^{\infty }\frac{1}{n!k!}\frac{1}{n+k}\frac{\left [ \Gamma \left ( n+k \right ) \right ]^{3}}{\Gamma \left ( 2n+2k \right )}$$
my attempt: $$I=2\sum_{k=1}^{\infty }\frac{\Gamma \left ( k \right )\Gamma \left ( \frac{1}{2} \right )}{k^{2}\Gamma \left ( k+\frac{1}{2} \right )}\,_3F_2\left ( k,k,k;k+1,k+\frac{1}{2};\frac{1}{4} \right )\left ( \frac{1}{4} \right )^{k}$$ but after this, how to deal with this hypergeometric function? any idea?
By using the integral representation of $_3F_2$ \begin{align*} _3F_2\left ( a,b,c;d,e;z \right )=&\frac{\Gamma \left ( d \right )\Gamma \left ( e \right )}{\Gamma \left ( a \right )\Gamma \left ( b \right )\Gamma \left ( d-a \right )\Gamma \left ( e-b \right )}\int_{0}^{1}\int_{0}^{1}x^{a-1}y^{b-1}\\ &\times \left ( 1-x \right )^{d-a-1}\left ( 1-y\right )^{e-b-1}\left ( 1-xyz \right )^{-c}\mathrm{d}x\mathrm{d}y \end{align*} we can perform the sum over $k$, \begin{align*} I&=2\sum_{k=1}^{\infty }\frac{\Gamma \left ( k \right )\Gamma \left ( \dfrac{1}{2} \right )}{k^{2}\Gamma \left ( k+\dfrac{1}{2} \right )}\,_3F_2\left ( k,k,k;k+1,k+\frac{1}{2};\frac{1}{4} \right )\left ( \frac{1}{4} \right )^{k}\\ &=2\sum_{k=1}^{\infty }\frac{1}{k}\int_{0}^{1}\int_{0}^{1}x^{k-1}y^{k-1}\left ( 1-y \right )^{-\frac{1}{2}}\left ( 1-\frac{1}{4}xy \right )^{-k}\left ( \frac{1}{4} \right )^{k}\mathrm{d}x\mathrm{d}y\\ &=2\int_{0}^{1}\int_{0}^{1}\frac{1}{xy\sqrt{1-y}}\sum_{k=1}^{\infty }\frac{1}{k}\left ( \frac{xy}{4-xy} \right )^{k}\mathrm{d}x\mathrm{d}y\\ &=-2\int_{0}^{1}\int_{0}^{1}\frac{1}{xy\sqrt{1-y}}\ln\left ( 1-\frac{xy}{4-xy} \right )\mathrm{d}x\mathrm{d}y\\ &=2\int_{0}^{1}\int_{0}^{1}\frac{1}{xy\sqrt{1-y}}\left [ \ln\left ( 1-\frac{xy}{4} \right )- \ln\left ( 1-\frac{xy}{2} \right )\right ]\mathrm{d}x\mathrm{d}y\\ &=-2\int_{0}^{1}\frac{1}{y\sqrt{1-y}}\mathrm{Li}_2\left (\frac{y}{4} \right )\mathrm{d}y+2\int_{0}^{1}\frac{1}{y\sqrt{1-y}}\mathrm{Li}_2\left (\frac{y}{2} \right )\mathrm{d}y \end{align*} For the first integral \begin{align*} -2\int_{0}^{1}\frac{1}{y\sqrt{1-y}}\mathrm{Li}_2\left (\frac{y}{4} \right )\mathrm{d}y&=-2\int_{0}^{1}\frac{1}{y\sqrt{1-y}}\sum_{n=1}^{\infty }\frac{y^{n}}{n^{2}}\left ( \frac{1}{4} \right )^{n}\mathrm{d}y\\ &=-2\sum_{n=1}^{\infty }\frac{1}{n^{2}}\left ( \frac{1}{4} \right )^{n}\int_{0}^{1}y^{n-1}\left ( 1-y \right )^{-\frac{1}{2}}\mathrm{d}y\\ &=-2\sum_{n=1}^{\infty }\frac{1}{n^{2}}\frac{\Gamma \left ( n \right )\Gamma \left ( \dfrac{1}{2} \right )}{\Gamma \left ( n+\dfrac{1}{2} \right )}\left ( \frac{1}{4} \right )^{n}\\ &=-_4F_3\left ( 1,1,1,1;\frac{3}{2},2,2;\frac{1}{4}\right )\\ &=\int_{0}^{1}\,_3F_2\left ( 1,1,1;\frac{3}{2},2;\frac{y}{4} \right )\mathrm{d}y \end{align*} In the last step we have used the integral representation of $_4F_3$ below $$_4F_3\left ( a,b,c,d;e,f,g;z\right )=\frac{\Gamma \left ( g \right )}{\Gamma \left ( g-d \right )}\int_{0}^{1}x^{d-1}\left ( 1-x \right )^{g-d-1}\,_3F_2\left (a,b,c;e,f;zx \right )\mathrm{d}x$$
By using the identity $(7.4.2.353)$ in Integral and series Vol.3 and afterwards making the substitution $\dfrac{\sqrt{y}}{2}=\sin\dfrac{u}{2}$ we get $$4\int_{0}^{1}\arcsin^{2}\frac{\sqrt{y}}{2}\frac{\mathrm{d}y}{y}=2\int_{0}^{\frac{\pi }{3}}\frac{u^{2}}{2\tan\dfrac{u}{2}}\, \mathrm{d}u$$ Integration by part we get $$2u^2\ln\left ( 2\sin\frac{u}{2} \right )\Bigg|_{0}^{\frac{\pi }{3}}-4\int_{0}^{\frac{\pi }{3}}u\ln\left ( 2\sin\frac{u}{2} \right )\mathrm{d}u=-4\int_{0}^{\frac{\pi }{3}}u\ln\left ( 2\sin\frac{u}{2} \right )\mathrm{d}u$$ Use the following identity in Polylogarithms and associated function $$\int_{0}^{\theta }\theta \ln\left ( 2\sin\frac{\theta }{2} \right )\mathrm{d}\theta =\mathrm{Li}_3\left ( 1 \right )-\theta \left [ \mathrm{Cl}_2\left ( \theta \right )-\frac{1}{\pi } \right ]\mathrm{Cl}_3\left ( \theta \right )$$ and $$\mathrm{Cl}_3\left ( \frac{\pi }{3} \right )=\frac{1}{3}\zeta \left ( 3 \right )$$ we get $$-2\int_{0}^{1}\frac{1}{y\sqrt{1-y}}\mathrm{Li}_2\left (\frac{y}{4} \right )\mathrm{d}y=\frac{8}{3}\zeta \left ( 3 \right )-\frac{4\pi }{3}\mathrm{Cl}_2\left ( \frac{\pi }{3} \right )$$ where $\mathrm{Cl}_n\left ( \cdot \right )$ is Clausen function.
Now using the same way we get \begin{align*} 2\int_{0}^{1}\frac{1}{y\sqrt{1-y}}\mathrm{Li}_2\left (\frac{y}{2} \right )\mathrm{d}y&=2\int_{0}^{1}\frac{1}{y\sqrt{1-y}}\sum_{n=1}^{\infty }\frac{y^{n}}{n^{2}}\left ( \frac{1}{2} \right )^{n}\mathrm{d}y\\ &=2\sum_{n=1}^{\infty }\frac{1}{n^{2}}\left ( \frac{1}{2} \right )^{n}\int_{0}^{1}y^{n-1}\left ( 1-y \right )^{-\frac{1}{2}}\mathrm{d}y\\ &=2\sum_{n=1}^{\infty }\frac{1}{n^{2}}\frac{\Gamma \left ( n \right )\Gamma \left ( \dfrac{1}{2} \right )}{\Gamma \left ( n+\dfrac{1}{2} \right )}\\ &=2\, _4F_3\left ( 1,1,1,1;\frac{3}{2},2,2;\frac{1}{2}\right )\\ &=2\int_{0}^{1}\,_3F_2\left ( 1,1,1;\frac{3}{2},2;\frac{y}{2} \right )\mathrm{d}y\\ &=4\int_{0}^{1}\arcsin^{2}\sqrt{\frac{y}{2}}\frac{\mathrm{d}y}{y}\\ &=2\int_{0}^{\frac{\pi }{2}}\frac{u^{2}}{2\tan\dfrac{u}{2}}\, \mathrm{d}u\\ &=\frac{\pi ^{2}\ln2}{4}-4\int_{0}^{\frac{\pi }{2}}u\ln\left ( 2\sin\frac{u}{2} \right )\mathrm{d}u\\ &=\frac{\pi ^{2}\ln2}{4}-\frac{35}{8}\zeta \left ( 3 \right )+2\pi \mathbf{G} \end{align*} where $\mathbf{G}$ is Catalan's constant.
Now, combine the two part of the result, we will get the final answer of the tough series $$\Large\boxed{\displaystyle \color{blue}{\sum_{n=0}^{\infty }\sum_{k=1}^{\infty }\frac{1}{n!k!}\frac{1}{n+k}\frac{\left [ \Gamma \left ( n+k \right ) \right ]^{3}}{\Gamma \left ( 2n+2k \right )}=-\frac{41}{24}\zeta \left ( 3 \right )-\frac{4\pi }{3}\mathrm{Cl}_2\left ( \frac{\pi }{3} \right )+\frac{\pi ^{2}\ln2}{4}+2\pi \mathbf{G}}}$$ Using the same idea, we can easily get $$\sum_{n=0}^{\infty }\sum_{k=1}^{\infty }\frac{\left ( -1\right )^{k}}{n!k!}\frac{1}{n+k}\frac{\left [ \Gamma \left ( n+k \right ) \right ]^{3}}{\Gamma \left ( 2n+2k \right )}=\frac{8}{3}\zeta \left ( 3 \right )-\frac{4\pi }{3}\mathrm{Cl}_2\left ( \frac{\pi }{3} \right )$$