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


2 Answers 2


This telescopes, using the fact that $\text{arctan}(u)-\text{arctan}(v) = \text{arctan}(\frac{u-v}{1+uv})$

Specifically take $u=n+1$ and $v=n$. Then $$\text{arctan}\left(\frac1{n^2+n+1}\right) = \text{arctan}(n+1)-\text{arctan}(n)$$

This gives $$\sum_{n=1}^{m}\arctan\left({\frac{1}{{n^2+n+1}}}\right) = \text{arctan}(m+1) - \pi/4$$


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.