2006 SMT/Advanced Topics Problems/Problem 10

Problem 10

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

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:

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