$\displaystyle \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{n^2 + n + 1}\right)$ equals

Observe that $\tan^{-1}\tfrac{1}{1 + n(n+1)} = \tan^{-1}\tfrac{(n+1)-n}{1+(n+1)n} = \tan^{-1}(n+1) - \tan^{-1}(n)$ (difference formula). **So the partial sum telescopes:** $\sum_{n=1}^{N} = \tan^{-1}(N+1) - \tan^{-1}(1)$. As $N \to \infty$, $\tan^{-1}(N+1) \to \tfrac{\pi}{2}$, giving $\tfrac{\pi}{2} - \tfrac{\pi}{4} = \tfrac{\pi}{4}$.