Question:
$\alpha$ is an irrational number.Denoted the fractional part of $n^2\alpha$ as $\{n^2\alpha\}$.Prove that $\{n^2\alpha\}$ is equidistributed,i.e. $$\lim_\limits{N\to\infty} \frac{\#\{1\leq n\leq N:\{n^2\alpha\}\in(a,b)\}}{N}=b-a,\forall (a,b)\subset [0,1).$$
Attempt:
By Weyl's criterion we just need to show that $$\frac1N \sum_\limits{n=1}^{N}e^{2\pi ikn^2\alpha}\to 0,\forall k\in \mathbb{N_+}.$$
I try to estimate the sum by integration,and get $\int_1^N e^{2\pi ikx^2\alpha} {\rm{d}}x=O(1)$.
But I have trouble estimating the difference $\sum_\limits{n=1}^{N-1}\int_n^{n+1} (e^{2\pi ikx^2\alpha}-e^{2\pi ikn^2\alpha}) {\rm{d}}x.$