Asymptotic behaviour of the number of sign changes in the sequence $\cos n\alpha$ Let $0$ $\leq$ $\alpha$ $\leq$ $\pi$. Denote by $V_n$$(\alpha)$ the number of sign changes in the sequence ${u_n}$ where $u_n$ $=$ $\displaystyle \cos n\alpha$. Then find the limit of the sequence $\frac{V_n(\alpha)}{n}$ as $n$ $\to$ $\infty$. 

I am not getting any idea as to how to proceed because the sign changes is irregular. Any help is appreciated.
The limit indeed exists and is
$$
\lim_{n\to\infty}\frac{V_n(\alpha)}n=\frac{\alpha}\pi.
$$
Assume first that $\alpha/\pi$ is rational, thus $\alpha=\pi p/q$ with $0\leqslant p\leqslant q$ and $q\geqslant1$. After $2q$ steps one is back at the angle $0$ on the circle and one did exactly $p$ full circles hence one experienced $2p$ changes of sign and the result holds.

(Here the number of changes of sign is the maximal length of a subsequence $(u_{n_k})$ such that $u_{n_k}u_{n_{k+1}}\lt0$ for every $k$.)

When $a$ is irrational, the rotation of angle $\alpha$ is ergodic with invariant measure the uniform measure on the circle hence the proportion of times one observes a change of sign converges to the combined measure of the portions $(\frac\pi2-\alpha,\frac\pi2)$ and $(-\frac\pi2-\alpha,-\frac\pi2)$, that is, to $2\alpha/(2\pi)$.
The number of signs is either even ($V_n = 2n$) or odd ($V_n = 2n+1$), and in both cases the answer is $2$.