2006 SMT/Advanced Topics Problems/Problem 10

Solution

First of all, remember that

$\arctan(a)+\arctan(b) = \arctan(\frac{a+b}{1-ab}) +k\pi$ Therefore we want $a$ and $b$ such that:

$1-ab = n^2-n+1 \Rightarrow ab = n(1-n)$ Noticing that $n+(1-n) = 1$ , we can set $a=n$ , and $b=(1-n)$ . Substituting this into the formula from the beginning, we see that:

$\sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^2-n+1}\right) =\sum_{n=1}^{\infty}\arctan(n)+\arctan(1-n)$

Using the identity that $\tan(x) = -\tan(-x)$

$\sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^2-n+1}\right) =\sum_{n=1}^{\infty}\arctan(n)-\arctan(n-1)$

This sum telescopes, leaving us with the first and last terms only; $-\arctan(0)$ which equals $0$ , and $\arctan(\infty) = \frac{\pi}{2}$ .

So our answer is:

$\sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^2-n+1}\right)= \boxed{\frac{\pi}{2}}$

Page

Toolbox

Search

2006 SMT/Advanced Topics Problems/Problem 10

Solution