How to prove that $n^{-2}[x+g(x)+g\circ g(x)+\cdots +g^{\circ n}(x)]$ converges when $n\to\infty$ Let $f:\mathbb R\to\mathbb R$ be a periodic function with period $1$. We assume that $f$ is Lipschitz continuous, and in particular, we assume that there exists an $L\in (0,1)$, such
that
$$
|f(x)-f(y)| \le L|x-y|, \quad \text{for all $x,y\in\mathbb R$.}
$$
Let also $g(x)=x+f(x).$
Show that the limit
$$\lim_{n\to +\infty}\frac1{n^2}[x+g(x)+g(g(x))+\cdots +g^{\circ n}(x)]$$
exists and is independent of $x$, where, for every $n\ge1$, $g^{\circ n}$ is $g$ composed with itself $n$ times, thus $g^{\circ 1}=g$ and, for every $n\ge1$, $g^{\circ n+1}=g\circ g^{\circ n}$.
My attempt: Since
$$f(x+1)=f(x),\forall x\in R$$
then
$$g(g(x))=g(x+f(x))=x+f(x)+f(x+f(x))$$
$$g(g(g(x)))=g(x+f(x)+f(x+f(x)))=x+f(x)+f(x+f(x))+f(x+f(x)+f(x+f(x)))$$
$$\cdots\cdots $$
and  I have
$$|g(x)-g(y)|=|x-y+f(x)-f(y)|\le |x-y|+|f(x)-f(y)|<(L+1)|x-y|$$
where $L+1>1$.
So $g$ is also Lipschitz continuous, and that's all I can do.
Thank you for you help.

First attempt. One needs to use the Stolz–Cesàro theorem. According to this theorem:
\begin{align}
\lim_{n\to \infty}\dfrac{x+g(x)+g(g(x))+\cdots +\underbrace{g(g(\cdots(g(x))))}_{n-1}}{n^2}&=\lim_{n\to \infty}\frac{\underbrace{g(g(\cdots(g(x))))}_{n}}{2n+1}\\=&\lim_{n\to \infty}\frac{\underbrace{g(g(\cdots(g(x))))}_{n+1}-\underbrace{g(g(\cdots(g(x))))}_{n}}{2},
\end{align}
provided that the last limit exists. Define $x_n=\underbrace{g(g(\cdots(g(x))))}_{n}$. Then
$$
\underbrace{g(g(\cdots(g(x))))}_{n+1}=g(x_n)=f(x_n)+x_n,
$$
and hence
$$
\underbrace{g(g(\cdots(g(x))))}_{n+1}-\underbrace{g(g(\cdots(g(x))))}_{n}=f(x_n)=f\bigg(\underbrace{g(g(\cdots(g(x))))}_{n}\bigg)
$$
We need show that the limit of $f(x_n)$ exists.
