Putnam 1986 
Topic: Calculus Trigonometry

Problem: Evaluate $\sum_{n=0}^\infty \mathrm{Arccot}(n^2+n+1)$, where $\mathrm{Arccot}\,t$ for $t \geq 0$ denotes the number $\theta$ in the interval $0 < \theta \leq \pi/2$ with $\cot \theta = t$.

Solution: Using \[ \cot(\alpha - \beta) = \frac{\cot \alpha \cot \beta + 1}{\cot \beta - \cot \alpha}, \] one sees that \[ \mathrm{Arccot}(1 + n + n^2) = \mathrm{Arccot} \, n - \mathrm{Arccot} (n + 1). \] Then the series telescopes to \[ \lim_{n \to \infty} (\mathrm{Arccot} \, 0 - \mathrm{Arccot} (n + 1)) = \boxed{\frac{\pi}{2}}. \]