Difference between revisions of "2006 GCTM State Tournament Problems/Individual Problem 46"
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− | <math>\ | + | <math>\cot^{-1} ( k^2 + k + 1 ) = \cot^{-1} ( k ) - \cot^{-1} ( k + 1 )</math> |
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− | <math>\sum_{k=0}^{n} \ | + | <math>\sum_{k=0}^{n} \cot^{-1} ( k^2 + k + 1 ) = \sum_{k=0}^{n} [ \cot^{-1} ( k ) - \cot^{-1} ( k + 1 )]</math>, |
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− | which is a [[telescoping series]] equal to <math>\ | + | which is a [[telescoping series]] equal to <math>\cot^{-1} 0 - \cot^{-1} ( n + 1 )</math>. We note that <math>\lim_{n\rightarrow \infty} \cot^{-1} ( n + 1 ) = 0</math>, so our infinite sum is equal to <math>\cot^{-1} 0</math>, which is equal to <math>\frac{\pi}{2}</math>. Q.E.D. |
[[Category:Intermediate Trigonometry Problems]] | [[Category:Intermediate Trigonometry Problems]] |
Revision as of 21:51, 10 January 2023
Problem
Find the exact value of the infinite series .
Solution
Motivated by the formula for the subtraction of cotangents, we observe that
Thus
,
which is a telescoping series equal to . We note that
, so our infinite sum is equal to
, which is equal to
. Q.E.D.