Why $\cos(n^2x) \not\to 0$ for any real $x$? I'd like to show, as simply as possible, that $\sum_{n=1}^\infty\cos(n^2 x)$ diverges for every real $x$.

(I know how to prove it for rational $x$. For irrational $x$, I don't know if $(n^2 x)$ is equidistributed mod 1. But if it is, perhaps this fact is not needed for a simple proof.)

Edit: Wikipedia says that Weyl proved in 1916 that $(n^2 x)$ is equidistributed mod 1 for irrational $x$, so this answers my question. The question remains if there's a simple proof not using this fact.
Assume that this sequence is convergent to $0$ so the subsequence $(\cos(4n^2x))$ is also convergent to $0$. Moreover by the identity

$$\cos^2 y+\sin^2y=1$$

the sequence $(\sin^2(n^2x))$ is convergent to $1$ but in this case  

$$\cos(4n^2x)=\cos^2(2n^2x)-\sin^2(2n^2x)\\=(\cos^2(n^2x)-\sin^2(n^2x))^2-4\sin^2(n^2x)\cos^2(n^2x)\xrightarrow{n\to\infty}1$$
contradiction!
A necessary condition for convergence of series $\displaystyle \sum_{1}^{\infty}a_n\,\,\, is: \lim_{n \to \infty} a_n = 0$.

The series $\sum_{n=1}^\infty\cos(n^2 x)$ not met for any x --> the series diverge.