TITLE: How to show the divergence of $\sum\limits_{n=1}^\infty\frac{\sin(\sqrt{n})}{\sqrt{n}}$
QUESTION [15 upvotes]: The 10 standard tests taught in class are:
1) $n^{th}$ term test for divergence.(Not applicable: $\lim =0$).
2) Geometric Series(Not applicable).
3) Telescoping Series(Not applicable)
4) Integral Test(Not applicable: $f<0$ sometimes)
5) $p$-series(Not applicable)
6) Direct Comparison(maybe)
7) Limit Comparison(Not applicable $a_n<0$ sometimes)
8) Alternating Series Test(Not Alternating)
9) Ratio Test fails
10) Root Test fails
I did find a hint online that states we should show that for $k^2+1\leq n\leq k^2+k$ we have $\sum\limits_{n=k^2+1}^{k^2+k}\frac{\sin(\sqrt{n})}{\sqrt{n}}>\frac{1}{8}$. Is there an easier way and if not how should we go about showing this?

REPLY [2 votes]: The hint you stated can't be true for all $k$ but it gives an idea on how to show the serie is divergent.
First remember that on $[2k\pi+\pi/4;2k\pi+3\pi/4]$ (for $k $ an integer) $\sin(x)\geq \sqrt 2 /2$. Now the condition $\sqrt n \in [2k\pi+\pi/4;2k\pi+3\pi/4]$ is equivalent to $n\in [4k^2\pi^2+k\pi^2+\pi^2/16;4k^2\pi^2+3k\pi^2+9\pi^2/16]$, and this last interval has a length of $2k\pi^2+\pi^2/2$, which is greater than $18k+2$ (using the fact that $\pi\geq3)$. So this interval contains at least $18k$ integers. 
Thus we have 
$$\sum_{2k\pi+\pi/4\leq \sqrt n\leq2k\pi+3\pi/4}\frac{\sin(\sqrt n)}{\sqrt n}\geq \sum_{2k\pi+\pi/4\leq \sqrt n\leq2k\pi+3\pi/4}\frac{\sqrt2/2}{ {2k\pi+3\pi/4}}\geq 18k\frac{\sqrt2}{2\cdot ( {2k\pi+3\pi/4})}$$
which is greater than some (strictly positive) constant.
So the serie must be divergent.
I dont think there is an (significantly) easier way to prove this result though.