In the question, I have to prove if it converges (no need to calculate sum) or diverges. I know this is some kind of telescopic series but I can't seem to get a right series to direct compare or in other ways.
$\sum_{n=1}^{\infty} \left(\arctan\left(\frac{2}{n}\right) – \arctan\left(\frac{1}{n}\right)\right)$.
Best Answer
With $f_n(x) = \arctan\left(\frac{x}{n}\right)$ and the MVT we know there exists a $\xi_n \in (1,2)$ s.t. $$\arctan\left(\frac{2}{n}\right) - \arctan\left(\frac{1}{n}\right) = f_n(2) - f_n(1) = \frac{f_n(2) - f_n(1)}{2-1} = f_n'(\xi_n) = \frac{n}{n^2 + \xi_n^2}$$ hence for $n > 1$ $$\arctan\left(\frac{2}{n}\right) - \arctan\left(\frac{1}{n}\right) = \frac{n}{n^2 + \xi_n^2}\ge \frac{n}{n^2 + n^2} \ge \frac{1}{2n}$$ and so $$\sum_{n>1} \left(\arctan\left(\frac{2}{n}\right) - \arctan\left(\frac{1}{n}\right)\right) \ge \sum_{n>1} \frac{1}{2n} = +\infty$$