Difference between revisions of "2006 SMT/Advanced Topics Problems/Problem 10"
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==Problem 10== | ==Problem 10== | ||
Evaluate: <math> \sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^2-n+1}\right) </math> | Evaluate: <math> \sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^2-n+1}\right) </math> | ||
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==Solution== | ==Solution== |
Latest revision as of 18:30, 14 January 2020
Problem 10
Evaluate:
Solution
First of all, remember that
Therefore we want
and
such that:
Noticing that
, we can set
, and
. Substituting this into the formula from the beginning, we see that:
Using the identity that
This sum telescopes, leaving us with the first and last terms only; which equals
, and
.
So our answer is: