TITLE: Bounding exponential sum with square roots
QUESTION [10 upvotes]: 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 $\epsilon>0$.

REPLY [6 votes]: The answer given by Xiaoyu He uses the same arguments as usual van der Corput's theorem (see Theorem 2.2 from Graham, S. W. & Kolesnik, G. "van der Corput's method of exponential sums"):

Suppose that $f$ is a real valued function
  with two continuous derivatives on $I$. Suppose also that there is some $\lambda > 0 $ and some $a > 1$ such that
  $\lambda  < |f''(x)| < a\lambda$
  on $I$. Then
  $$\sum_{n\in I} e(f(n))\ll a|I|\lambda^{1/2} + \lambda^{-1/2}.$$

So we can try to apply this theorem directly. For $f(x)=\sqrt{ N x}$ on the interval $I(X)=[X,2X)$ we have $f''(x)\asymp\lambda =N^{1/2}X^{-3/2}$. Van der Corput's theorem gives the estimate
$$\sum_{n\in I(X)} e(f(n))\ll N^{1/4}X^{1/4} + N^{-1/4}X^{3/4}.$$
Summation over $X=1,2,4,8,\ldots,2^k\asymp N$ leads to the upper bound 
$$\sum_{n=1}^N e(f(n))\ll N^{1/2}.$$