Distribution of the sequence $\bigl (\frac{\phi(n)}{n}\bigr )_{n=1}^{\infty}$ in $[0,1]$ Maybe this is a well-know problem. 
What do we know about distribution of the sequence $\bigl (\frac{\phi(n)}{n}\bigr )_{n=1}^{\infty}$ in $[0,1]$? (Where $\phi$ is the Euler's totient function).
In the other words does there exist a distribution like $\mu$ concentrated on $[0,1]$ that for every interval $[a,b]$ we have the following equation?
$$\lim_{n\to \infty}  \frac{\# \ \{i \ |\ \frac{\phi(i)}{i} \in [a,b], \ 1\leq i \leq n \}}{n} = \mu([a,b]) $$
Is the sequence equidistributed in $[0,1]$?
 A: As Lucia points out the existence of a limiting distribution for $\phi(n)/n$ is well-known.
The most direct way to show that there is a limiting distribution is to compute the moment
$$
\sum_{n \leq x} \Big ( \frac{\phi(n)}{n} \Big )^{k}
$$
for all fixed positive integers $k$. The limiting distribution $G(t)$ is continuous but not differentiable. It is also known to be purely singular, so that $G'(t) = 0$ almost everywhere in $[0,1]$. In addition, for the modulus of continuity we know that
$$
\sup_{0 \leq t \leq 1} |G(t) - G(t - \varepsilon t)| \ll \frac{1}{\log(1/\varepsilon)}
$$
and that this is optimal.  Finally the behavior as $t \rightarrow 0$ and $t \rightarrow 1$ is markedly different. As $t \rightarrow 0$ there is a doubly exponential decay in $1/t$, while the behavior at $t \rightarrow 1$ is such that $1 - G(1 - 1/\sigma) \sim c/\log \sigma$ as $\sigma \rightarrow \infty$. 
For proofs and more informations a good starting point is 
http://iecl.univ-lorraine.fr/~Gerald.Tenenbaum/PUBLIC/Prepublications_et_publications/EulerLocal.pdf
http://arxiv.org/pdf/1011.4262v1.pdf
and the references there-in.
