Does the series $\;\sum\limits_{n=0}^{\infty}\left(\frac{\pi}{2} - \arctan(n)\right)$ converge or diverge? [closed]

Solution 1:

Hint: $$\sum_{n=0}^{\infty} \left(\frac{\pi}{2} - \arctan(n)\right) = \dfrac{\pi}{2} + \sum_{n=1}^{\infty} \left(\arctan\left(\frac 1n\right)\right)$$

Then clearly, as $\;n \to \infty,\; \dfrac 1n \to 0$

Perhaps use the limit comparison test? Perhaps the integral test?

Additional hint:

To use the limit comparison test, as suggested by Mhenni in the comment below, consider $\displaystyle \sum b_n = \sum \left(\dfrac 1n\right)\,,\;\,$ and note that $$\;\lim_{n \to \infty} \frac{\arctan(1/n)}{1/n}=1,\;$$ meaning the two series either converge together or diverge together. So apply what you know about the behavior of $\displaystyle \sum_{n=1}^\infty \dfrac 1n\;$ to the task at hand.

Solution 2:

The serie $\sum \arctan \frac{1}{n}$ is divergent for two reasons: first, it's a positive serie and second $\arctan\frac{1}{n}\sim \frac{1}{n}$ so by comparaison we conclude since the serie $\sum \frac{1}{n}$ is divergent.

To explain more why the positivity of the serie is important to use comparaison test, this is a counterexample : the serie $\sum \log(1+\frac{(-1)^n}{\sqrt{n}})$ is divergent while $\log(1+\frac{(-1)^n}{\sqrt{n}})\sim\frac{(-1)^n}{\sqrt{n}}$ and the serie $\sum \frac{(-1)^n}{\sqrt{n}}$ is convergent