TITLE: Convergence of $\sum^\infty_{n=1}\arctan(\frac 1 {\sqrt n}) $ and how to approach trigonometric expressions in sums
QUESTION [5 upvotes]: Does $$\sum^\infty_{n=1}\arctan\left(\frac 1 {\sqrt n}\right)$$ converge?
The series probably diverges and I should probably use the comparison test, but I don't know what to use.
Note: no integral test.
My other question is, in general, when there are trigonometric expressions in series, what is the recommended approach?

REPLY [2 votes]: This diverges by a simple comparison test.
$$
\sum^\infty_{n=1}\arctan\left(\frac 1 {\sqrt n}\right) \ge \sum_{n=1}^\infty \frac 1 {2\sqrt{n}} = \infty
$$
because $\arctan x\ge \dfrac x 2$ if $0\le x\le 1$.
So it doesn't have much to do with the behavior of trigonometric functions generally.