I was observing this interesting paper and conjecture this result, $ $ \sum_{n=0}^{\infty}\frac{{2n \choose n}}{2^{4n+1}}^2\cdot \frac{n(6n-1)}{(2n-1)^2(2n+1)}=\frac{C}{\pi} \tag1$ $ Where C is Catalan’s constant $ =0.9156965…$ I am unable to present a prove of $ (1)$ . How do we go about to prove its?Read more