Show that $\lim\limits_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n^2}$ exists and is independent of the choice of $a$ Suppose $f:\mathbb{R}\to\mathbb{R}$ has period 1, and for some $q\in(0,1)$:
$$|f(x)-f(y)|\leq q|x-y|\quad \forall x,y$$
Let $g(x)=x+f(x)$, for any $a\in\mathbb{R}$,define the following sequence:
$$a_1=a,\quad a_2=g(a_1),\quad a_3=g(a_2),\quad \dots,\quad a_{n+1}=g(a_n)$$
Show that
$$\lim_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n^2}$$
exists and does not depend on the choice of $a$.
I don't even know what is the possible approach, any hints? Thanks!
 A: This is not a complete solution. I hope someone can help to fully nail it down.
It will suffice using Stolz-Cesàro theorem (page 1) for $b_n=2n-1,$ (because $\sum_{i=1}^nb_i=n^2$) to show that $\lim_{n\to \infty}\dfrac{a_n}{2n-1} < \infty.$ Again employing the other form of the theorem (Remark in page 2), it will suffice to prove that $\lim_{n\to \infty}\dfrac{a_{n}-a_{n-1}}{(2n-1)-(2n-3)} < \infty$ which is to check $ \lim_{n\to \infty} f(a_n) < \infty.$
We know $f$ is Lipschitz, so is continuous. Therefore, it boils down to showing that $\lim_{n\to \infty} a_n < \infty.$ That is, proving that the iterated sequence of $g,$ i.e. $\{g^n(a)\}_{n \in \mathbb{N} }$ where $g^n(a)=g(g^{n-1}(a))$ is the $n$ times composition of $g$ with itself, has a limit. But the problem is that $g$ is a non-contracting Lipschitz, i.e. $|g(x)-g(y)| \leq (q+1)|x-y|$ where $1<q+1<2$ and this causes problems, of course. 
The fact is that I haven't used the periodic condition on $f$ yet, and so I think, this should help resolve the issue, but I couldn't figure out how!   
