Difference between revisions of "2006 GCTM State Tournament Problems/Individual Problem 46"

Latest revision as of 21:51, 10 January 2023

Problem

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

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.

@@ Line 1: / Line 1: @@
 == Problem ==
-Find the exact value of the infinite series <math>\sum_{k=0}^{\infty} \mbox{Cot}^{-1} ( k^2 + k + 1 )</math>.
+Find the exact value of the infinite series <math>\sum_{k=0}^{\infty} \cot^{-1} ( k^2 + k + 1 )</math>.
 == Solution ==

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Difference between revisions of "2006 GCTM State Tournament Problems/Individual Problem 46"

Latest revision as of 21:51, 10 January 2023

Problem

Solution