*The following result is new and is going to be published on Arxiv.org in the upcoming days with many more interesting results by Jacopo D' Aurizio who has actually proved it. Nevertheless , I am posting it here since it is interesting , challenging as well as approachable using only elementary tools.*

Prove that

\begin{align*}

{}_4 F_3 \left ( 1, 1, 1 , \frac{3}{2} ; \frac{5}{2} , \frac{5}{2} , \frac{5}{2} ; 1 \right ) &= 27 \sum_{n=0}^{\infty} \frac{16^n}{\left ( 2n+3 \right )^3 \left ( 2n+1 \right )^2 \binom{2n}{n}^2} \\

&= \frac{27}{2} \bigg( 7 \zeta(3) + \left ( 3 - 2 \mathcal{G} \right ) \pi - 12 \bigg)

\end{align*}

where $\mathcal{G}$ denotes the Catalan's constant.