In some recent work I found I needed to prove a central limit theorem for 
the interesting series:


$\sum\_{n=1}^\infty \cos (u \log p\_n) $


where u is a random variable uniform on the interval $[0,2\pi]$ and $p\_n$ is the n-th prime number. If the primes were truly random, this
series would essentially be like a random walk and the CLT definitely applies. The primes 
appear random enough but are not completely random. I checked numerically that the probability distribution for the above series is indeed gaussian (normal), so I believe it does indeed have a CLT. 


I'm aware that the CLT is known to apply to some series that are not independently identically distributed random variables if the correlations between them are very weak. The formalism for the latter involves the notion of $\alpha$-mixing which I do not know much about. The above series is similar to the lacunary trigonometric series, but fails the Hadamard gap condition, so those theorems (Salem-Zygmund) do not apply. 


Does anyone have any suggestions for how to approach this? Is there a simple way to see if the series would pass the $\alpha$-mixing criterion? I'm not sure I want to invest all the time to learn about the CLT for series with $\alpha$ mixing, so am looking for something simpler. 


(the recent work referred to relates this series to the Riemann hypothesis and is at math.NT if you are interested in how the series originated.)



A central limit theorem for a trigonometric series involving primes
If you only look inside $[0,2\pi]$, then it is not true that the distribution is Gaussian. 


EDIT


Set
$$
S(x;t) = \sum\_{p\le x} p^{it},
$$
so that the sum you are looking at is the real part of $S(x;t)$. Then the Prime Number Theorem implies that for $t\in[0,2\pi]$ we have that
$$
S(x;t) = \frac{x^{1+it}}{(1+it)\log x} + O\left(\frac{x}{\log^2x}\right).
$$
Therefore
$$
\mu:=\frac{1}{2\pi} \int\_0^{2\pi} S(x;t)dt \ll \frac{x}{\log^2x},
$$
by integration by parts, and
$$
\sigma^2:= \frac{1}{2\pi} \int\_0^{2\pi}|S(x;t)|^2 dt \sim \frac{cx^2}{\log^2x},
$$
for some $c>0$. Consequently,
$$
\begin{align}
M\_k&:=\frac{1}{2\pi} \int\_0^{2\pi}\left| \frac{S(x;t)-\mu}{\sigma}\right|^{2k} dt \\
 & = \frac{1}{2\pi} \int\_0^{2\pi}\left| \frac{S(x;t)(1+O(1/\log x))}{\sigma}\right|^{2k} dt \\
 &\sim \frac{1}{2\pi c^{k}}\int\_0^{2\pi} \frac{dt}{(1+t^2)^k} 
 \in[c\_1^k,c\_2^k]
\end{align}
$$
for all fixed $k\in\mathbb{N}$, where $c\_1$ and $c\_2$ are certain positive constants. If the distribution of $S(x;t)$ were Gaussian, then $M\_k$ would grow like $c\_0^k k!$ for some $c\_0>0$, much faster than its current growth.


END OF EDIT


You do get a Gaussian distribution, but you have to look at appropriate ranges of $t$ with respect to $x$ (and we have to look at long intervals of $t$, because we don't understand well what happens inside very short intervals). This is part of the context of the Selberg Central Limit Theorem for the Riemann zeta function. In the proof of his theorem, Selberg shows that the statistical behavior of $\log\zeta(1/2+it)$ for $t\in[T,2T]$ can be modeled by the statistical behavior of $\sum\_{p\le T^\epsilon} 1/p^{1/2+it}$ for some $\epsilon=\epsilon(T)$ that tends to 0 slowly. Then he proceeds to estimate moments of the latter sum (which is essentially equivalent to studying the distribution of the sum you're interested in).


The rough idea is that
$$
\begin{align}
\int\_T^{2T} S(x;t)^k \overline{S(x;t)}^{\ell} dt
 &= \sum\_{p\_1,\dots,p\_{k+\ell}\le x} \int\_T^{2T} \left(\frac{p\_1\cdots p\_k}{p\_{k+1}\cdots p\_{k+\ell}}\right)^{it} dt \\
 &= T\sum\_{\substack{ p\_1,\dots,p\_{k+\ell}\le x \\ p\_1\cdots p\_k=p\_{k+1}\cdots p\_{k+\ell}}} 1
+ O\left( \sum\_{\substack{ p\_1,\dots,p\_{k+\ell}\le x \\ p\_1\cdots p\_k\neq p\_{k+1}\cdots p\_{k+\ell}}} \frac{1}{|\log(p\_1\cdots p\_k/(p\_{k+1}\cdots p\_{k+\ell}))|}\right).
\end{align}
$$
The main term above is 0 unless $k=\ell$, in which case it equals asymptotically $T$ times the $2k$-th moment of a Gaussian. The error term can be shown to be small if $x^{\max\{k,\ell\}}\le T^{1/4}$. Indeed, in this case $p\_1\cdots p\_k,p\_{k+1}\cdots p\_{k+\ell}\le T^{1/4}$ and thus $|\log(p\_1\cdots p\_k/(p\_{k+1}\cdots p\_{k+\ell}))|\ge 1/T^{1/4}$. So the total error term is $T^{1/4} \pi(x)^{k+\ell} \le T^{3/4}$, which is small enough for our purposes. Hence, if $x=T^{o(1)}$, then an increasingly (as $T\to\infty$) large number of moments is shown to match the moments of the Gaussian, and standard probability results then imply that the distribution of $S(x;t)$ for $t\in[T,2T]$ is Gaussian.


