Bound for sum of fractional parts of $\sqrt{k}\, \alpha$ Let us consider the sum
$$C_\alpha(n)=\sum_{1\leq k\leq n} \left(\{ k\alpha\}-\frac{1}{2}\right)$$
where $\alpha$ is a positive irrational real number and $\{ x \}$ denotes the fractional part of $x$. The term $k \alpha$ is uniformly distributed modulo $1$. As reminded in this paper, Sierpinski showed that 
$$C_\alpha(n)=o(n)$$
and stronger bounds were successively shown.
Now let us consider the sum
$$C_\alpha(n, 1/2)=\sum_{1\leq k\leq n} \left(\{ \sqrt{k} \,\alpha\}-\frac{1}{2}\right)$$
The term $(\sqrt{k}\, \alpha)$ is also uniformly distributed modulo $1$, and an $O(\sqrt{n})$ bound could be hypothesized, as cited in some comments of this previous question. I wonder whether this bound (or some stronger ones) can be formally proven, and whether there are references for this. 
 A: First let us show that no stronger bound than $O(\sqrt{n})$ can hold. Consider a large $m$, and put
$$N_1 = N_1(m) = \biggl\lceil\biggl(\frac{m}{\alpha}\biggr)^2\biggr\rceil,\qquad N_2 = N_2(m) = \biggl\lfloor\biggl(\frac{m + \frac{1}{3}}{\alpha}\biggr)^2\biggr\rfloor\,.$$
Then $N_2 - N_1 \approx \frac{2m}{3\alpha^2} \approx \frac{2}{3\alpha}\sqrt{N_1} \approx \frac{2}{3\alpha}\sqrt{N_2}$, and for $N_1 \leqslant k \leqslant N_2$ we have $0 \leqslant \alpha\sqrt{k} \leqslant \frac{1}{3}$, so
$$C_{\alpha}(N_2,1/2) - C_{\alpha}(N_1,1/2) \leqslant -\frac{1}{6}(N_2 - N_1) \approx -\frac{1}{9\alpha}\sqrt{N_j}$$
for $j \in \{1,2\}$. Hence at least one of $C_{\alpha}(N_1(m),1/2) > \frac{1}{30\alpha}\sqrt{N_1(m)}$ and $C_{\alpha}(N_2(m),1/2) < -\frac{1}{30\alpha}\sqrt{N_2(m)}$ holds for all sufficiently large $m$.
To show the $O(\sqrt{n})$ bound, put $M = \lfloor \alpha \sqrt{n}\rfloor$. There are $O(\sqrt{n})$ values $k$ for which $\alpha\sqrt{k} \geqslant M$, the corresponding terms hence contribute at most $O(\sqrt{n})$. For $1 \leqslant m \leqslant M$, the sum over those $k$ for which $m-1 \leqslant \alpha\sqrt{k} < m$ is bounded by a constant (depending on $\alpha$), and there are $M = O(\sqrt{n})$ such sums, hence we have an overall $O(\sqrt{n})$ bound.
It remains to show that the sums mentioned above are indeed uniformly bounded. For $m \geqslant \alpha + 1$, put
$$x = \biggl(\frac{m-1}{\alpha}\biggr)^2,\qquad y = \biggl(\frac{m}{\alpha}\biggr)^2\,.$$
The sum we want to estimate is
$$\sum_{k = \lceil x\rceil}^{\lceil y\rceil - 1} f(k)\,,$$
where $f(t) = \alpha \sqrt{t} - \bigl(m - \frac{1}{2}\bigr)$. Since $f$ is (strictly) increasing, we have
\begin{align}
\sum_{k = \lceil x\rceil}^{\lceil y\rceil - 1} f(k)
&< \int_{\lceil x\rceil}^{\lceil y\rceil} f(t)\,dt \\
&= \int_x^y f(t)\,dt + \int_{y}^{\lceil y\rceil} f(t)\,dt - \int_x^{\lceil x\rceil} f(t)\,dt
\end{align}
and
\begin{align}
\sum_{k = \lceil x\rceil}^{\lceil y\rceil - 1} f(k)
&> \int_{\lceil x\rceil - 1}^{\lceil y\rceil - 1} f(t)\,dt \\
&= \int_{x}^{y} f(t)\,dt + \int_{\lceil x\rceil - 1}^{x} f(t)\,dt - \int_{\lceil y\rceil - 1}^y f(t)\,dt\,.
\end{align}
Also $\lvert f(t)\rvert \leqslant \frac{1}{2}$ for $x \leqslant t \leqslant y$, and $f(y+1) - f(y) = \alpha (\sqrt{y+1} - \sqrt{y}) = \frac{\alpha}{\sqrt{y+1} + \sqrt{y}} < \frac{\alpha}{2\sqrt{y}} = \frac{\alpha^2}{2m} \leqslant \frac{1}{2}\alpha^2$, $f(x) - f(x-1) = \frac{\alpha}{\sqrt{x} + \sqrt{x-1}} \leqslant \frac{\alpha}{\sqrt{x}} = \frac{\alpha^2}{m-1} \leqslant \alpha^2$. Thus $\lvert f(t)\rvert \leqslant \frac{1}{2} + \alpha^2$ for $\lceil x\rceil - 1 \leqslant t \leqslant \lceil y\rceil$ and
$$\biggl\lvert \sum_{k = \lceil x\rceil}^{\lceil y\rceil - 1} f(k) - \int_x^y f(t)\,dt\biggr\rvert < 1 + 2\alpha^2\,.$$
The integral is easily evaluated,
$$\int_x^y f(t)\,dt = \frac{2\alpha}{3}\bigl(y^{3/2} - x^{3/2}\bigr) - \bigl(m - \tfrac{1}{2}\bigr)(y-x) = \frac{2m^2 - 2m + \frac{2}{3}}{\alpha^2} - \frac{2m^2-2m + \frac{1}{2}}{\alpha^2} = \frac{1}{6\alpha^2}\,,$$
whence
$$\biggl\lvert \sum_{k = \lceil x\rceil}^{\lceil y\rceil - 1} f(k)\biggr\rvert \leqslant 1 + 2\alpha^2 + \frac{1}{6\alpha^2}$$
for $m \geqslant \alpha + 1$.
For $m < \alpha + 1$, i.e. $m \leqslant \lceil \alpha\rceil$ there is of course a common bound (these are $\lceil\alpha\rceil$ fixed sums, independent of everything except $\alpha$), say $B_{\alpha}$, so $K_{\alpha} := \max\:\{B_{\alpha}, 1 + 2\alpha^2 + 1/(6\alpha^2)\}$ is a bound holding uniformly for all $m$.
Note that irrationality of $\alpha$ played no role here, only $\alpha > 0$ was used (and for $\alpha < 0$ little would change).
A: Assuming $k^{1/2}\alpha$ is never an integer
$$\begin{eqnarray}\sum_{k\le n} \{k^{1/2}\alpha\} &=& \sum_{k\le n}k^{1/2}\alpha-\sum_{k\le n}\lfloor k^{1/2}\alpha \rfloor=\sum_{k\le n}k^{1/2}\alpha- \sum_{k\le n,m< k^{1/2}\alpha} 1\\&=&\sum_{k\le n}k^{1/2}\alpha-\sum_{m\le n^{1/2}\alpha}\ \sum_{k\le n,k^{1/2}\alpha>m} 1\\ &=&\sum_{k\le n}k^{1/2}\alpha-\sum_{m\le n^{1/2}\alpha}(n-\lfloor m^2/\alpha^2\rfloor)\\&=&\sum_{k\le n}k^{1/2}\alpha-\sum_{m\le n^{1/2}\alpha}(n-m^2/\alpha^2)+O(n^{1/2})\end{eqnarray}$$
The remaining part is routine
