Difference between revisions of "2006 SMT/Advanced Topics Problems/Problem 10"
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| + | ==Problem 10== | ||
| + | Evaluate: <math> \sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^2-n+1}\right) </math> | ||
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| + | [[2006 SMT/Advanced Topics Problems/Problem 10|Solution]] | ||
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==Solution== | ==Solution== | ||
Revision as of 18:30, 14 January 2020
Problem 10
Evaluate: 
Solution
First of all, remember that
![\[\arctan(a)+\arctan(b) = \arctan(\frac{a+b}{1-ab}) +k\pi\]](http://latex.artofproblemsolving.com/c/8/5/c852e16348119665b4d182721fcb3a9f0537f193.png)
Therefore we want 
and 
such that:
![\[1-ab = n^2-n+1 \Rightarrow ab = n(1-n)\]](http://latex.artofproblemsolving.com/d/9/7/d97a318e78d7dd73efb61928849c19e001079335.png)
Noticing that 
, we can set 
, and 
. Substituting this into the formula from the beginning, we see that:
![\[\sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^2-n+1}\right) =\sum_{n=1}^{\infty}\arctan(n)+\arctan(1-n)\]](http://latex.artofproblemsolving.com/6/e/f/6ef6896be34f90210f5ac26ae273f13fb02a0a1f.png)
Using the identity that 
![\[\sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^2-n+1}\right) =\sum_{n=1}^{\infty}\arctan(n)-\arctan(n-1)\]](http://latex.artofproblemsolving.com/f/a/4/fa468b3ebee0c960a26ecfc3955593d164f34c1c.png)
This sum telescopes, leaving us with the first and last terms only; 
which equals 
, and 
.
So our answer is:
![\[\sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^2-n+1}\right)= \boxed{\frac{\pi}{2}}\]](http://latex.artofproblemsolving.com/5/f/5/5f545be8318eb3625eaff0a6946ba0660e5292c1.png)
