How to show the divergence of ∞∑n=1sin(√n)√n\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?

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).