Prove that $\lim\limits_{n\to\infty} a_n$ exists. 
Let $a_1,\cdots, a_{2022}$ be real numbers in $(0,1)$. Define for $n\ge 2023$, $a_n = (a_{n-1}+\cdots + a_{n-2022})^{1/2023}$. Prove that $\lim\limits_{n\to\infty} a_n$ exists.

Clearly it suffices to show that $\lim\limits_{n\to\infty} a_n^{2023}$ exists since $x\mapsto x^{1/2023}$ is a continuous function (on $\mathbb{R}$). I initially thought of using the Cesaro stolz theorem, but it seems hard to prove that the limit $\lim\limits_{n\to\infty} \dfrac{a_{n+1}^{2023}-a_n^{2023}}{1} = \lim\limits_{n\to\infty} a_n - a_{n-2022}$ exists. But  it should be possible to prove it indeed exists and is equal to zero. The Squeeze theorem or some sort of telescoping sum might help. I'm not sure if inequalities like the AM-GM inequality could be useful for this problem.
 A: Let $S_n=a_{n-1}+\cdots+a_{n-2022}$ for $n\ge 2023$.
First, notice that $a_n>0$ for all $n$, shown by induction.
Assume that $a_n<1$ for all $n$. Then for $n\ge 2023$ we have $a_n>a_n^{2023}=S_n>\max_{i=1,\dots,2022}a_{n-i}$, so with induction we obtain that $a_n$ is strictly increasing for $n\ge 2023$, hence it converges.
Otherwise, let $n_\circ$ be the smallest integer with $a_{n_\circ}\ge 1$, in particular $n_\circ\ge 2023$.
With induction we get $S_n>1$ and hence $a_n>1$ for $n\ge n_\circ$.
Let $n_*=n_\circ+2022$. For $n_\circ\le n<n_*$ we have
$a_{n}>1>a_{n-2022}$, so $S_{n+1}-S_{n}=a_{n}-a_{n-2022}>0$ and further $a_{n+1}=S_{n+1}^{1/2023}>S_{n}^{1/2023}=a_{n}$, which gives $a_{n}>a_k$ for $k<n$ and $n_\circ\le n\le n_*$. The same induction gives $a_n>a_k$ for $k<n$ and $n_\circ\le n$, since we have $a_n>a_{n-2022}$ in the induction step directly from the induction hypothesis. Hence, the sequence is eventually strictly increasing and converges.
Notice that in any case the sequence is strictly increasing for $n\ge 2023$, as pointed out by Stephen in the comment.
In order to identify fixed points of the recursion consider the solutions of the equation $x=(2022x)^{1/2023}$ for $x>0$, which are exactly the roots of
$f(x)=x-(2022x)^{1/2023}=x^{1/2023}(x^{2022/2023}-2022^{1/2023})$. Thus, we clearly have a unique root at $a_*=2022^{1/2022}>1$.
Since the initial elements are smaller than $1$ and $a_*>1$, we use induction to obtain $a_n<(2022 a_*)^{1/2023}=a_*$ for all $n$.
In particular, this yields $a_\circ=\lim_{n\rightarrow\infty}a_n\le a_*$. Now, for arbitrarily small $\varepsilon$ there exists $n_\circ$ such that $a_\circ-\varepsilon< a_n< a_\circ$ for all $n\ge n_\circ$. Hence, for $n>n_\circ+2022$ we have $a_\circ> a_n>(2022(a_\circ-\varepsilon))^{1/2023}$. Taking $\varepsilon$ to $0$ gives $a_\circ\ge(2022a_\circ)^{1/2023}$ since the right hand side is continuous, i.e. $f(a_\circ)\ge 0$. But the product form of $f$ above immediately yields $f(x)<0$ for $0<x<a_*$ since $x^{2022/2023}$ is increasing. With $a_\circ\le a_*$ this yields $a_\circ=a_*$.
So, the sequences $a_n$ converges to $a_*$ independent of the initial conditions.
