Probabilistic convergence of $\cos n\Theta$ with $\Theta$ uniform on $[0,2\pi]$ This is a homework question so I'd appreciate advice rather than a solution.
I'm asked to determine in what modes the following sequence of random variables converges. In particular, I am interested in convergence in probability (measure) and also a bit in almost sure convergence.
$$X_n=\cos n\Theta$$
where $\Theta$ is uniform on the interval $[0,2\pi]$.
My attempt:
First a few thoughts: It feels a bit like convergence in probability here is related to the rationals being dense in the reals. Whereas a.s. convergence does not hold since the rationals are only a countable subset and have measure $0$. Well, this is the gut feeling I had anyway.
Focusing on convergence in probability (measure), rewriting
$$X_n= \cos (n \Theta \mod \pi)$$
it is enough to show that $n\Theta$ converges $\mod \pi$. I.e. we want to show that
$$P(|n \Theta - m\pi|>\epsilon)\to 0$$
for some $m \in \mathbb{N}$.
I don't really see where to go from here or how to show that the above statement is true (maybe it isn't?). Am I missing something?
Any advice is much appreciated.
 A: EDIT3: Sorry, I haven't looked for a while. I solved the problem a few days back and was preoccupied with deadlines. 
One notes that the distribution is invariant under $n$ and hence converges to an inverse trigonometric in distribution (in fact the same for all $n$).
I looked at this the wrong way. It is easy to show that $X_n$ is not Cauchy in mean square. Consider
$$
E[X_n X_m]=\left\{\begin{array}{ccc}
0 & if & n \neq m, \\
 \frac{1}{2\pi}\int_{0}^{2\pi} \cos^2(nx) dx\neq 0 & if & n=m.
\end{array}\right.
$$
Since $E[|X_n-X_m|^2]=E[X_n^2]+E[X_m^2]-2E[X_nX_m]$ the limit as $n,m \to \infty$ depends on the relation of $n,m$ so the sequence is not Cauchy. Hence it does not converge in $L^2$. 
Next it can be shown that $X_n \to X$ in $L^2$ is equivalent to $X_n \to X$ in probability whenever $|X_n|\leq C$, $C>0$ almost surely for all $n\in \mathbb{N}$. This clearly holds for $C=1$ in this case hence we can conclude that the sequence does not converge in probability either (nor can it converge almost surely since this is stronger than in probability).
A: For the distribution part: I think that for $P(cos(n\Theta)\leq t)=P(\Theta\leq \frac{cos^{-1}(t))}{n})$. See what this yields and take it to $\infty$
A: Your calculation of the cumulative distribution function is wrong.
Consider the cumulative distribution function for $X = \cos(\Theta)$ and $X = \cos(2 \Theta)$. Do they differ? What would be the c.d.f for $X = \cos(n \Theta)$? 
Now proving that the sequence does not converge in $L^2$ but is bounded almost surely is enough to state that it does not converge in probability or almost surely. 
