A geometry problem by Akshat Sharda

Note that $\tan^{-1}(n+1) - \tan^{-1}n \; = \; \tan^{-1}\left(\frac{(n+1)-n}{1 + n(n+1)}\right) \; = \; \cot^{-1}\big(1 + n(n+1)\big)$ so that $\sum_{k=1}^\infty \cot^{-1}\left(1 + 2\sqrt{\sum_{r=1}^k r^3}\right) \; = \; \sum_{k=1}^\infty \big[\tan^{-1}(k+1) - \tan^{-1}k\big] \; =\; \lim_{K \to \infty}\tan^{-1}(K+1) - \tan^{-1}1 \; = \; \tfrac14\pi$ and so the sum is $\sum_{n=1}^\infty 4^{-n} \; = \; \tfrac13$ making the answer $\boxed{3}$ .