It is well known that for each $m\in\mathbb{N}$
$$\lim\_{N\to\infty}\frac1N\sum\_{n=1}^Ne^{2\pi i\sqrt{nm}}=0$$
My question is whether there is some uniformity in the variable $m$.
More precisely, is it true that
$$\lim\_{N\to\infty}\frac1N\sum\_{n=1}^Ne^{2\pi i\sqrt{nN}}=0?$$
I would be particularly interested in a power saving of the form
$$\lim\_{N\to\infty}\frac1{N^{1-\epsilon}}\sum\_{n=1}^Ne^{2\pi i\sqrt{nN}}=0$$
for some $\epsilon>0$.



Bounding exponential sum with square roots
I write $e(x)=e^{2\pi i x}$. Here is a naive bound via [Kuzmin's estimate](http://www.math.harvard.edu/~elkies/M259.02/kmv.pdf), which says that if you have a sequence $c\_n$ with monotonic differences $\delta\_n$ all in some interval $[k+\epsilon, k+1-\epsilon]$, then the entire exponential sum is small:


$$|\sum\_n e(c\_n)| \ll \epsilon^{-1}$$


If we write $c\_n = \sqrt{nN}$ then in Elkies' notation 


$$\delta\_n = c\_n-c\_{n-1} = \frac{\sqrt{N}}{\sqrt{n} + \sqrt{n-1}}$$


Now pick an $\epsilon>0$ and break up the interval $[1,N]$ into two pieces - those numbers where the fractional part of $\delta\_n$ is within $\epsilon$ of an integer, and everything else. The possible integers go up to $O(\sqrt{N})$, and for each $m \ll \sqrt{N}$ it is easy to check that the number of $|\delta\_n - m| < \epsilon$ is $\ll \epsilon N/m^2$. Cutting off the interval $[1, \epsilon N]$ to be safe, we can say that the total area of these exceptional intervals is $O(\epsilon N)$.


On the other hand the rest of the $c\_n$ are broken up into $O(\sqrt{N})$ intervals with sum at most $\epsilon^{-1}$ each. Thus,


$$|\sum\_n e(c\_n)| \ll \epsilon N + \epsilon^{-1} \sqrt{N}.$$


Picking $\epsilon = N^{-1/4}$, this should give a power-saving, for any $\delta>0$:


$$\lim\_{N\rightarrow\infty} \frac{1}{N^{\frac{3}{4}+\delta}}\sum\_{n\le N} e(\sqrt{nN}) \rightarrow 0.$$


