TITLE: Explicitly finding the sum of $\arctan(1/(n^2+n+1))$
QUESTION [17 upvotes]: This is from a GRE prep book, so I know the solution and process but I thought it was an interesting question: Explicitly evaluate $$\sum_{n=1}^{m}\arctan\left({\frac{1}{{n^2+n+1}}}\right).$$

REPLY [7 votes]: By observation, 
$\tan^{-1}\frac{1}{n^2+n+1}=\cot^{-1}(n^2+n+1)=\cot^{-1}\frac{n(n+1)+1}{n+1-n}$
$=\cot^{-1}(n)-\cot^{-1}(n+1)$
$\sum_{n=1}^{m}\tan^{-1}\left({\frac{1}{{n^2+n+1}}}\right)$
$=\frac{\pi}{4}-\cot^{-1}(m+1)$ using this.