Let $N$ be a positive integer. The sequence $x_1, x_2, \ldots$ of non-negative reals is defined by $$x_n^2=\sum_{i=1}^{n-1} \sqrt{x_ix_{n-i}}$$ for all positive integers $n>N$. We aim to prove that : $$x_{n}=\frac{\pi}{8}n+ o(n)$$
For this let $y_{n}:= \frac{x_{n}}{n}$. We have that :$$y_{n}^{2}= \frac{1}{n} \sum_{i=1}^{n-1} \sqrt{y_{i}\cdot y_{n-i}} \sqrt{\frac{i}{n} \cdot \frac{n-i}{n}}~~(1)$$
We can notice from here that due to Rieman sum :$$\frac{1}{n} \sum_{i=1}^{n-1} \sqrt{\frac{i}{n} \cdot \frac{n-i}{n}} ~~\to_{n \to +\infty} \int_{0}^{1} \sqrt{x\cdot (1-x)} = \frac{\pi}{8}$$
Now we are left to prove that:
- $(y_{n})$ converges
- Taking the limit in $(1)$ lead to :$$\lim~~ y_{n} = \lim ~~\frac{1}{n} \sum_{i=1}^{n-1} \sqrt{\frac{i}{n} \cdot \frac{n-i}{n}}=\frac{\pi}{8} $$
For this let $u_{n}:= \frac{1}{n} \sum_{i=1}^{n-1} \sqrt{\frac{i}{n} \cdot \frac{n-i}{n}}$ and $l=\lim u_{n}$. Consider $r_{n}=\frac{y_{n}}{l}$ we aim to prove that $\lim ~~r_{n}=1$.
For this we have from $(1)$ :
$$u_{n}\cdot \min\{y_{k} : k< n\} \leq y_{n}^{2} \leq u_{n} \cdot \max\{y_{k}: k < n\}$$
But here I can find ways to conclude. Any help or reference if such exercice is standard would be welcome. Thanks.
