2006 GCTM State Tournament Problems/Individual Problem 46

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Problem

Find the exact value of the infinite series $\sum_{k=0}^{\infty} \mbox{Cot}^{-1} ( k^2 + k + 1 )$.

Solution

Motivated by the formula for the subtraction of cotangents, we observe that

$\displaystyle \mbox{Cot}^{-1} ( k^2 + k + 1 ) = \mbox{Cot}^{-1} ( k ) - \mbox{Cot}^{-1} ( k + 1 )$

Thus

$\sum_{k=0}^{n} \mbox{Cot}^{-1} ( k^2 + k + 1 ) = \sum_{k=0}^{n} [ \mbox{Cot}^{-1} ( k ) - \mbox{Cot}^{-1} ( k + 1 )]$,

which is a telescoping series equal to $\displaystyle \mbox{Cot}^{-1} 0 - \mbox{Cot}^{-1} ( n + 1 )$. We note that $\lim_{n\rightarrow \infty} \mbox{Cot}^{-1} ( n + 1 ) = 0$, so our infinite sum is equal to $\displaystyle \mbox{Cot}^{-1} 0$, which is equal to $\displaystyle \frac{\pi}{2}$. Q.E.D.