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.
Best Answer
Let's look at the beginning of the sequence, where $n \leq 2022.$ Due to monotonicity, we have that $a_n < 1$ implies $a_n^{2022/2023} < 1,$ and because $a_n > 0$ implies $a_n^{1/2023} > 0,$ multiplying on both sides we have that $a_n < a_n^{1/2023}$ for all $n \leq 2022.$ We must also have that $a_n < \sum_{k = 1}^{2022} a_k,$ due to $a_n > 0$ for $n \leq 2022,$ so $$a_n < a_n^{1/2023} < \left(\sum_{k = 1}^{2022} a_k\right)^{1/2023} = a_{2023}$$
Now note that when we calculate $a_{2024},$ this new value $a_{2023}$ will replace $a_1$ in the sum inside the power, leading to an even larger result. This leads us to the following statement:
We will prove this by strong induction: our statement above shows that this is true for the base case $n = 2023.$
Now supposing that we have $a_n > a_{n-1}$ for all $2023 \leq n < k,$ we want to prove it true for $n = k.$
It will help to show that $a_{k-1} > a_{k - 2023}$: if $k \geq 4045$ then $k - 2023 \geq 2022$ and the desired inequality drops out from simply repeatedly applying the inequality $a_n > a_{n-1}$. Otherwise, for $2023 < k < 4045$ we still have $a_{k-1} \geq a_{2023},$ and because $k - 2023 < 2022$ we have $a_{2023} > a_{k - 2023}$ by our earlier work, so $a_{k-1} > a_{k - 2023}.$
Recall that by definition we have $$a_k = \left(\sum_{m = k-2022}^{k-1} a_m\right)^{1/2023}, \ \ \ a_{k-1} = \left(\sum_{m = k-2023}^{k-2} a_m\right)^{1/2023}$$ Rewriting inside the power in the first equation we get $$a_k = \left(\sum_{m = k-2023}^{k-2} a_m + (a_{k-1} - a_{k-2023})\right)^{1/2023}$$
and because $a_{k-1} > a_{k-2023}, a_{k-1} - a_{k-2023} > 0 $ and consequently $$\sum_{m = k-2023}^{k-2} a_m + (a_{k-1} - a_{k-2023}) > \sum_{m = k-2023}^{k-2} a_m$$
and again by the monotonicity of the power function here this yields
$$\left(\sum_{m = k-2023}^{k-2} a_m + (a_{k-1} - a_{k-2023})\right)^{1/2023} > \left(\sum_{m = k-2023}^{k-2} a_m\right)^{1/2023} \\ \therefore a_k > a_{k-1}$$
concluding the proof.
This nearly concludes the proof that $a_n$ converges. All that remains is to show that $a$ is bounded above, which is rather simple: let's choose $a_n < 2.$
It's clear that this bound holds for $n \leq 2022,$ since we have $a_n < 1$ there. For $n \geq 2023,$ we will again continue using strong induction: suppose that $a_n < 2$ is true for all $n < k.$ Naturally,
$$\sum_{m = k-2022}^{k-1} a_m < \sum_{m = k-2022}^{k-1} 2 = 4044 < 2^{2023} \\ a_k = \left(\sum_{m = k-2022}^{k-1} a_m\right)^{1/2023} < \left(2^{2023}\right)^{1/2023} = 2$$
proving that $a_n < 2$ for all natural numbers $n,$ and because $a_n$ is both monotonically increasing and bounded above, it must converge.
I hope this helps. Sorry if any of the writing got a bit gnarly, I've tried to keep everything as readable as possible but it can be tough with all the $2022$s and $2023$s going around. (My apologies in advance as well for any minor errors, there were a few picky details I had to go back and forth on a bit so I could have missed something.)