8. Evaluate $\sum_{n=0}^{\infty} \cot ^{-1}\left(n^{2}+n+1\right)$.
Solution: $\sum_{n=0}^{\infty} \cot \left(n^{2}+n+1\right)=\sum_{n=0}^{\infty} \arctan \left(\frac{1}{n^{2}+n+1}\right)=\sum_{n=0}^{\infty} \arctan (n+1)-\arctan (n)$ by the sum/difference formula for tangent. This sum, taken out to $n=N$, telescopes to $-\arctan (0)+\arctan (N+1)$. So as $N$ goes to infinity, the sum goes to $\pi / 2$.