By using partial summation and Weyl's inequality, it is not hard to show that the series $\sum_{n\geq 1}\frac{\sin(n^2)}{n}$ is convergent.
I recall a generalization of partial summation formula:
Suppose that $\lambda_1,\lambda_2,\ldots$ is a nondecreasing sequence of real numbers with limit infinity, that $c_1,c_2,\ldots$ is an arbitrary sequence of real or complex numbers, and that $f(x)$ has a continuous derivative for $x\geq \lambda_1$. Put $$ C(x)=\sum_{\lambda_n\leq x}c_n, $$ where the summation is over all $n$ for which $\lambda_n\leq x$. Then for $x\geq\lambda_1$, $$ \sum_{\lambda_n\leq x}c_nf(\lambda_n)=C(x)f(x)-\int^{x}_{\lambda_1}C(t)f'(t)dt.\tag 1 $$
Now we can write if $y=x^2$ and $\lambda_n=n^2$ and $C(t)=[\sqrt{t}]$ (integer part of $\sqrt{t}$): $$ S=\sum_{1\leq n\leq x}\frac{\sin(n^2)}{n^a}=\sum_{\lambda_n\leq y}\frac{\sin(\lambda_n)}{\lambda_n^{a/2}}= $$ $$ =[\sqrt{y}]\frac{\sin(y)}{y^{a/2}}-\int^{y}_{1}[\sqrt{t}]\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt. $$ But it is $[\sqrt{t}]=\sqrt{t}-\{\sqrt{t}\}$, where $\{\sqrt{t}\}$ is the fractional part of $\sqrt{t}$. Hence $$ S=-\frac{1}{2}Re\left[iy^{1/2-a/2}E\left(\frac{1+a}{2},iy\right)\right]+\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]+\sin(1)-\{\sqrt{y}\}\frac{\sin(y)}{y^{a/2}}+ $$ $$ +\int^{y}_{1}\{\sqrt{t}\}\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt, $$ where $$ E(a,z)=\int^{\infty}_{1}\frac{e^{-tz}}{t^a}dt $$ But when $a>0$ and $y\rightarrow+\infty$ we have $$ \lim_{y\rightarrow+\infty}\left\{-\frac{1}{2}Re\left[iy^{1/2-a/2}E\left(\frac{1+a}{2},iy\right)\right]+\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]\right\}+\sin(1)= $$ $$ =\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]+\sin(1) $$ Also $x$ is positive integer and $\{\sqrt{y}\}=0$.
Hence when $a>0$, then $$ \lim_{x\rightarrow\infty}\sum^{x}_{n=1}\frac{\sin(n^2)}{n^a}=\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]+\sin(1)+\lim_{y\rightarrow\infty}\int^{y}_{1}\{\sqrt{t}\}\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt $$ But $$ \int^{y}_{1}\{\sqrt{t}\}\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt=\int^{y}_{1}\{\sqrt{t}\}\frac{\cos(t)t^{a/2}-a/2\sin(t)t^{a/2-1}}{t^a}dt= $$ $$ \int^{y}_{1}\{\sqrt{t}\}\left(\cos(t)t^{-a/2}-a/2\sin(t)t^{-a/2-1}\right)dt=\int^{y}_{1}\{\sqrt{t}\}\frac{\cos(t)}{t^{a/2}}dt-\frac{a}{2}\int^{y}_{1}\frac{\sin(t)}{t^{a/2+1}}\{\sqrt{t}\}dt. $$ Clearly when $a$ is positive and constant $$ \lim_{y\rightarrow+\infty}\int^{y}_{1}\frac{\sin(t)}{t^{a/2+1}}\{\sqrt{t}\}dt=2\lim_{x\rightarrow\infty}\int^{x}_{1}\frac{\sin(t^2)}{t^{a+1}}\{t\}dt<\infty, $$ since $0\leq\{t\}<1$ and $-1\leq\sin(t^2)\leq 1$, for all $t>0$.
Hence it remains to find under what condition on $a>0$ we have $$ \int^{\infty}_{1}\{\sqrt{t}\}\frac{\cos(t)}{t^{a/2}}dt=2\int^{\infty}_{1}\cos(t^2)t^{1-a}\{t\}dt<\infty, $$ knowinig already that for all $0<a\leq 1$ we have $$ \int^{\infty}_{1}\cos(t^2)t^{1-a}dt<\infty. $$
However it is $$ F(x)=\int^{x}_{1}\cos(t)\{\sqrt{t}\}dt=\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2}{\pi}}\right)-\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2x}{\pi}}\right)+ $$ $$ +\sum_{2\leq k<\sqrt{x}}\sin(k^2)+\{\sqrt{x}\}\sin(x)=O\left(\sum_{2\leq k<\sqrt{x}}\sin(k^2)\right).\tag 2 $$ The function $\textrm{Fs}(x)$ is the FresnelS function $$ \textrm{Fs}(z):=\int^{z}_{0}\sin\left(\pi t^2/2\right)dt $$ and it is known that $$ \lim_{x\rightarrow+\infty}\textrm{Fs}(x)=\frac{1}{2}. $$ Hence if we set $$ S(x):=\sum_{2\leq k<x}\sin(k^2) $$ and assume that $a=1/2-\epsilon$, $\epsilon>0$, then $$ I(x)=\int^{x}_{1}\frac{1}{t^{a/2}}\cos(t)\{\sqrt{t}\}dt=\int^{x}_{1}\frac{F'(t)}{t^{a/2}}dt=\frac{F(x)}{x^{a/2}}+\frac{a}{2}\int^{x}_{1}\frac{F(t)}{t^{a/2+1}}dt=S_1+S_2,\tag 3 $$ where $$ S_1=\frac{1}{x^{a/2}}S(\sqrt{x})+\frac{\{\sqrt{x}\}\sin x}{x^{a/2}}+\frac{\sqrt{\pi/2}}{x^{a/2}}\left(\textrm{Fs}\left(\sqrt{\frac{2}{\pi}}\right)-\textrm{Fs}\left(\sqrt{\frac{2x}{\pi}}\right)\right) $$ and $$ S_2=\frac{a}{2}\int^{x}_{1}\frac{1}{t^{1/4-\epsilon/2+1}}\{\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2}{\pi}}\right)-\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2t}{\pi}}\right)+ $$ $$ +\{\sqrt{t}\}\sin(t) +\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\}dt. $$ But it is known that there exists constant $C$ such that for infinite values of $x\in\textbf{N}$ holds $$ \sum_{2\leq k\leq x}\sin(k^2)>Cx^{1/2}.\tag 4 $$ Hence for infinite values of $x$ we will have (easily) $$ S_1>C_1x^{\epsilon/2}.\tag 5 $$ Moreover if we assume that $$ \left|\sum_{2\leq k\leq x}\sin(k^2)\right|=O\left(x^{c+\delta}\right)\textrm{, }\forall \delta>0\textrm{ and }x\rightarrow+\infty,\tag 6 $$ then in view of (4) it must be $c\geq 1/2$. Also $$ S_2=C_{\epsilon}(x)+\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)dt. $$ Hence $$ \left|S_2\right|=\left|C_{\epsilon}(x)+\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)dt\right|\leq $$ $$ \leq \left|\left|C_{\epsilon}(x)\right|+\left|\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)dt\right|\right|\leq $$ $$ |C_{\epsilon}(x)|+\int^{x}_{1}\left|t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)\right|dt= $$ $$ =\left|C_{\epsilon}(x)\right|+C_2\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left|\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right|dt\leq $$ $$ \leq\left|C_{\epsilon}(x)\right|+C_2\int^{x}_1t^{-1-1/4+\epsilon/2}t^{1/4+\delta/2}dt= $$ $$ =\left|C_{\epsilon}(x)\right|+C_2\int^{x}_{1}t^{-1+\epsilon/2+\delta/2}dt=\left|C_{\epsilon}(x)\right|+\frac{2}{\delta+\epsilon}\left(x^{(\delta+\epsilon)/2}-1\right)< $$ $$ <|C_{0}|+\log x+C_3d\log^2 x,\tag 7 $$ where $\epsilon>0$ and $\delta>0$ so small as we please and $d=\frac{\epsilon+\delta}{2}>0$, $C_3>0$ constant. It is also easy to see someone that $\left|C_{\epsilon}(x)\right|$ are bounded by a constant $C_0>0$.
Hence from $(3)$ and $(5),(7)$ we have if $a=1/2-\epsilon$, that $$ \left|\int^{x}_{1}\frac{\cos (t)\{\sqrt{t}\}}{t^{a/2}}dt\right|=|S_1+S_2|\geq |S_1|-|S_2|>C_1x^{\epsilon/2}-|C_0|-\log x-C_3d\log^2 x, $$ For infinite values of $x\in\textbf{N}$.
Hence $$ \lim_{x\rightarrow+\infty}\int^{x}_{1}\frac{\cos(t)\{\sqrt{t}\}}{t^{a/2}}dt=+\infty $$ and we conclude that if (6) holds, then $\textrm{inf}\geq1/2$.
I will argue now about the the case $a=\frac{1}{2}+2\epsilon$, $\epsilon>0$ i.e the case when $a$ is not $1/2$ but rather a limiting case and doesnot cover the case $a=1/2$. Both results $\textrm{inf}\geq1/2$ and $\textrm{inf}=1/2+2\epsilon$, clearly show us that for $1/2<a\leq1$ the sum $\sum^{\infty}_{n=1}\frac{\sin(n^2)}{n^a}$ converges and diverges for $0<a<1/2$, under the hypothesis $(6)$. For $a=1/2$, we dont know.
For $a=1/2+2\epsilon$, $\epsilon>0$ and for $x>>1$, we chose $\delta>0$ such that $$ S\left(\sqrt{x}\right)\leq C_1x^{1/4+\delta}, $$ we get $$ \left|I(x)\right|=|S_1+S_2|\leq \left|C_1\frac{S\left(\sqrt{x}\right)}{x^{a/2}}+C_2\frac{a}{2}\int^{x}_{1}\frac{S\left(\sqrt{t}\right)}{t^{a/2+1}}dt\right|\leq $$ $$ \leq\left|C_1'\frac{x^{1/4+\delta}}{x^{1/4+\epsilon}}+C_2'\left(\frac{1}{4}+\epsilon\right)\int^{x}_{1}\frac{t^{1/4+\delta}}{t^{1+1/4+\epsilon}}dt\right|= $$ $$ =\left|C_1'x^{-(\epsilon-\delta)}+C_2'\left(\frac{1}{4}+\epsilon\right)\int^{x}_{1}\frac{dt}{t^{1+\epsilon-\delta}}\right|. $$ For $\delta=\epsilon/2$ we get $$ |I(x)|\leq \left|C_1'x^{-\epsilon/2}-\frac{2C_2'}{\epsilon}\left(\frac{1}{4}+\epsilon\right)\left(x^{-\epsilon/2}-1\right)\right|= $$ $$ =\left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)+2C_2'\left(1-x^{-\epsilon/2}\right)+H-H\right|\leq $$ $$ \leq\left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)+2C_2'\left(1-x^{-\epsilon/2}\right)+H\right|+\left|H\right|.\tag 8 $$ Now we set $$ X=C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)>0 $$ and $$ Y=2C_2'\left(1-x^{-\epsilon/2}\right)+H>0 $$ and I use the inequality $$ \left|X+Y\right|\leq \left|X-\frac{Y}{4\epsilon}\right|,\tag 9 $$ which is is true for small $\epsilon$ and $x>1$ since we can write equivalent $$ \left|X+Y\right|^2\leq \left|X-\frac{Y}{4\epsilon }\right|^2\Leftrightarrow X^2+Y^2+2XY\leq X^2+\frac{Y^2}{16\epsilon^2}-\frac{XY}{2\epsilon }\Leftrightarrow $$ $$ Y^2+2XY\leq\frac{Y^2}{16\epsilon^2}-\frac{XY}{2\epsilon}\Leftrightarrow Y+2X\leq \frac{Y}{16\epsilon^2}-\frac{X}{2\epsilon}\Leftrightarrow $$ $$ \left(\frac{1}{16\epsilon^2}-1\right)Y\geq X\left(2+\frac{1}{2\epsilon}\right) $$ This last inequality holds for all small $\epsilon>0$ and $x>>1$ since it can be writen equivalently as $$ (1-16\epsilon^2)Y-X(32\epsilon^2+8\epsilon)\geq0\Leftrightarrow $$ $$ 2\epsilon(1+2\epsilon)\left(C_2'-4\epsilon C_1'+4C_2'\epsilon\right)x^{-\epsilon/2}\geq 0, $$ where we have used the value $$ H=\frac{2(C_2'+4C_2\epsilon)}{1-4\epsilon} $$ Hence (9) is true and we can extract from relation (8) the conclusion $$ \left|I(x)\right|\leq \left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)+2C_2'\left(1-x^{-\epsilon/2}\right)+H\right|+\left|H\right|= $$ $$ =\left|X+Y\right|+\left|H\right|\leq $$ $$ \leq\left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)-\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)-\frac{H}{4\epsilon}\right|+|H|= $$ $$ =\left|C_1'x^{-\epsilon/2}-\frac{H}{4\epsilon}\right|+|H|. $$ Hence $$ \epsilon |I(x)|\leq C_1'x^{-\epsilon/2}\epsilon+H/4+|H|\epsilon. $$ Hence we conclude that $$ \epsilon \left|I(x)\right|=O(1) $$ is bounded. Hence for $\epsilon>0$ small but constant the $I(x)$ are bounded.
