2006 GCTM State Tournament Problems/Individual Problem 46

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

$\cot^{-1} ( k^2 + k + 1 ) = \cot^{-1} ( k ) - \cot^{-1} ( k + 1 )$

Thus

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

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

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2006 GCTM State Tournament Problems/Individual Problem 46

Problem

Solution