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

 
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<math>\displaystyle \mbox{Cot}^{-1} ( k^2 + k + 1 ) = \mbox{Cot}^{-1} ( k ) - \mbox{Cot}^{-1} ( k + 1 )</math>
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<math>\cot^{-1} ( k^2 + k + 1 ) = \cot^{-1} ( k ) - \cot^{-1} ( k + 1 )</math>
 
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<math>\sum_{k=0}^{n} \mbox{Cot}^{-1} ( k^2 + k + 1 ) = \sum_{k=0}^{n} [ \mbox{Cot}^{-1} ( k ) - \mbox{Cot}^{-1} ( k + 1 )]</math>,  
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<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>\displaystyle \mbox{Cot}^{-1} 0 - \mbox{Cot}^{-1} ( n + 1 )</math>.  We note that <math>\lim_{n\rightarrow \infty} \mbox{Cot}^{-1} ( n + 1 ) = 0</math>, so our infinite sum is equal to <math> \displaystyle \mbox{Cot}^{-1} 0</math>, which is equal to <math>\displaystyle \frac{\pi}{2}</math>.  Q.E.D.
+
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 20:51, 10 January 2023

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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