Convergence $ \sum \frac{\sin \sqrt{n}}{n^{3/2}}, \quad \sum \frac{\sin n}{\sqrt{n}}, \quad\sum \frac{\sin \sqrt{n}}{n^{3/4}} $ I need to examine the convergence of the following infinite series:
$$ \sum_{n=1}^\infty \frac{\sin \sqrt{n}}{n^{3/2}},  \,\,\,\, \sum_{n=1}^\infty \frac{\sin n}{\sqrt{n}}, \,\,\,\,\sum_{n=1}^\infty \frac{\sin \sqrt{n}}{n^{3/4}} $$
I was able to show that the first two converge. For (1) I used  $|\frac{\sin n}{n^{3/2}}| \leq \frac{1}{n^{3/2}}$ and the comparison  with a convergent p-series  to prove it converges.
For (2) I used the Dirichlet test knowing that $\sum_{n=1}^m \sin n$ is bounded for all $m$ to show it converges.
With the third series, the comparison $|\frac{\sin \sqrt{n}}{n^{3/4}}| \leq \frac{1}{n^{3/4}}$ does not help and I’m pretty sure $\sum_{n=1}^m \sin \sqrt{n}$  is not bounded.
I am unsure how to make progress with this third series.
 A: The third series is convergent.  While not monotonic, a type of integral test applies.
We can show in general that a series $\sum_{k=1}^\infty f(k)$ and integral $\int_1^\infty f(x) \, dx$ converge and diverge together if $f'$ is absolutely integrable over $[1,\infty).$ 
The convergence of the series
$$\sum_{k=1}^\infty \frac{\sin \sqrt{k}}{k^{3/4}},$$
follows from the convergence of the improper integral
$$\int_1^\infty \frac{\sin \sqrt{x}}{x^{3/4}} \, dx = \int_{1}^{\infty}\frac{2u\sin u}{u^{3/2}} \, du = 2 \int_{1}^{\infty}\frac{\sin u}{\sqrt{u}} \, du,$$
where the RHS integral converges by the Dirichlet test.
Integrating by parts, we see
$$\int_{k-1}^k (x - \lfloor x\rfloor)f'(x) \, dx= \int_{k-1}^k (x - k +1)f'(x) \, dx \\ = \left.(x - k +1)f(x)\right|_{k-1}^k - \int_{k-1}^k f(x) \, dx \\ = f(k) - \int_{k-1}^k f(x) \, dx.$$
Hence,
$$|C_k| := \left|f(k) - \int_{k-1}^kf(x) \, dx\right| \leqslant \int_{k-1}^k\left| (x - \lfloor x \rfloor )f'(x)\right| \, dx \leqslant \int_{k-1}^k|f'(x)| \, dx$$
and we have absolute convergence
$$\sum_{k=2}^\infty|C_k| \leqslant \int_1^\infty |f'(x)| \, dx < \infty.$$
This implies convergence of 
$$\sum_{k=2}^\infty C_k = \sum_{k=2}^\infty f(k) - \int_1^\infty f(x) \, dx,$$
whence the series and integral must converge or diverge together.
In this case with $f(x) = \sin \sqrt{x}/x^{3/4}$ we have
$$\int_1^\infty |f'(x)| \, dx  = \int_1^\infty \left|\frac{\cos \sqrt{x}}{2x^{5/4}} - \frac{3 \sin \sqrt{x}}{4 x^{7/4}} \right| \, dx < \infty$$
