TITLE: Recurrence Relation involving Fractional Part
QUESTION [7 upvotes]: I am considering the following sequence:
$$a_0=\sqrt 2$$
$$a_{n+1}=a_n+\{a_n\}$$
where $\{x\}:= x-\lfloor x\rfloor$ denotes the fractional part function.
Since I have observed that the sequence exhibits almost-linear growth, I am trying to find the value of the limit
$$\lim_{n\to\infty} \frac{a_n}{n}$$
This is by no means rigorous, but I believe that the value is $1/2$, because the $\{a_n\}$ seems to behave somewhat like a random variable, and if we instead considered the sequence
$$b_{n+1}=b_n+X_n$$
where each $X_n$ is a random variable uniformly distributed in $(0,1)$, the expected value of $\Delta b_n=X_n$ is $1/2$.
Any ideas about how to prove this more rigorously? Is my reasoning even correct?

REPLY [4 votes]: Somewhat rigorous...
Let $\lfloor a_n \rfloor = I_n$, $\{a_n\} = f_n$.The problem can be decomposed into 
\begin{align}
f_{n+1} &= 2f_n - \mathbb{1}_{f_n \geq \frac{1}{2}} \\
I_{n+1} &= I_n + \mathbb{1}_{f_n \geq \frac{1}{2}}
\end{align}
The fractional part follows a Bernoulli map, whose invariant density is the uniform distribution if $f_0$ is irrational, so that $$ \lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^n\mathbb{1}_{f_k \geq \frac{1}{2}} = \frac{1}{2}.$$
Therefore $$\lim_{n\to \infty} \frac{a_n}{n}= \lim_{n\to \infty} \frac{I_n}{n} =\lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^n\mathbb{1}_{f_k \geq \frac{1}{2}} = \frac{1}{2}.$$