Convergence of $\sum^\infty_{n=1}\arctan(\frac 1 {\sqrt n}) $ and how to approach trigonometric expressions in sums 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?
 A: 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
$$
\sum_{n=1}^\infty \arctan\left(\frac{1}{\sqrt{n}}\right) \geq \frac{1}{2}\sum_{n=1}^\infty \frac{1}{\sqrt{n}} = +\infty.
$$
A: We have
$$\lim_{n\to +\infty}\sqrt{n}\arctan\frac{1}{\sqrt{n}} = 1, $$
hence the series
$$\sum_{n\geq 1}\arctan\frac{1}{\sqrt{n}}$$
is divergent by asymptotic comparison with the series
$$ \sum_{n\geq 1}\frac{1}{\sqrt{n}}$$
that is divergent.
For a slightly different approach, notice that $\arctan x$ is a concave function on the interval $I=[0,1]$, hence for every $n\geq 1$ we have: 
$$\arctan\frac{1}{\sqrt{n}}\geq\frac{\pi}{4\sqrt{n}}\geq\frac{\pi}{2}\left(\sqrt{n+1}-\sqrt{n}\right)$$
so that:
$$ \sum_{n=1}^{N}\arctan\frac{1}{\sqrt{n}}\geq \frac{\pi}{2}\left(\sqrt{N+1}-1\right).$$
A: You have, when $x \to 0$,$$
\arctan x=x+O(x^3)
$$ which gives, when $n \to \infty$, $$
\arctan(\frac 1 {\sqrt n})=\frac 1 {\sqrt n}+O(\frac 1 {n^{3/2}})
$$ and by comparison the series is divergent.
A: 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.
