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?
Best Answer
Here is a non-asymptotic solution.
Notice that for all $x \leq 1$, $$ \arctan(x) = \int_0^x \frac{dt}{1+t^2} \geq \int_0^x\frac{dt}{2} = \frac{x}{2}. $$ Therefore $$ \sum_{n=1}^\infty \arctan\left(\frac{1}{\sqrt{n}}\right) \geq \frac{1}{2}\sum_{n=1}^\infty \frac{1}{\sqrt{n}} = +\infty. $$