How to show the divergence of $\sum\limits_{n=1}^\infty\frac{\sin(\sqrt{n})}{\sqrt{n}}$ 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? 
 A: Hint: if you can find $k$ so that $\sqrt{k^2+1}$ is really close to $\dfrac{\pi}{2}$, then for $i$ at most $k$, $\sqrt{k^2+i}$ will also be close to $\dfrac{\pi}{2}$ (Because the difference between $\sqrt{k^2+1}$ and $\sqrt{k^2+i}$ is at most $0.5$, since $(\sqrt{k^2+1}+\dfrac{1}{2})^2>k^2+k$. 
Now for such $k$, $\sin(\sqrt{k^2+i})$ for $i$ at most $k$ will be bounded below by some constant $C$(they found something that works with 1/8, but you can find one of your own). 
Then:
$$\sum_{n=k^2+1}^{n=k^2+k} \dfrac{\sin\sqrt n}{\sqrt n} > C \sum_{n=k^2+1}^{n=k^2+k} \dfrac{1}{\sqrt n} > C \sum_{n=k^2+1}^{n=k^2+k} \frac{1}{k} = C$$
A: 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.
