Difficult Asymptotic of a Recursive Sequence Let $x_1>1$ and $x_n$ defind by iteration by
$$x_{n+1}=x_n+\frac{1}{\prod_{k=1}^nx_k^{1/n}}$$

Prove that there exists $C>0$ such that $x_n \sim C\sqrt{n}$.

This exercise seems difficult. Indeed, here is my start.
First, remark that $x_n$ is positive, hence increasing. Therefore, $x_{n+1}-x_n \geqslant \frac{1}{x_n}$, therefore $x_{n+1}^2-x_n^2 \geqslant 2$, and we obtain $x_n \geqslant C_1 \sqrt{n}$. Plugging this bound we obtain similarly that $x_n \leqslant C_2\sqrt{n}$ and finally we have $x_n = O(\sqrt{n})$ which is a good start.
Next, we want to prove that $x_n / \sqrt{n}$ indeed converge. In order to do that we juste have to prove that $x_{n+1}^2-x_n^2$ is a convergent sequence which I can not prove. This is equivalent to proving that $\frac{x_n}{\prod_{k=1}^nx_k^{1/n}}$ converge.
Here are the details of a further attempt: as we expect $y_n:=x_n/\sqrt{n}$ to converge, we re-write everything in terms of $y_n$ and obtain:
$$y_{n+1}^2-y_n^2=-\frac{y_n^2}{n+1} + \frac{2\sqrt{n}y_n}{(n+1)(n!)^{\frac{1}{2n}}\prod y_k^{1/n}}=\frac{y_n}{n+1} \left(\frac{2\sqrt{n}}{(n!)^{\frac{1}{2n}}\prod y_k^{1/n}}-y_n\right)$$
and now, because of Stirling, we expect that $\lim y_n^2 = \lim y_n \prod y_k^{1/n} = 2\sqrt{e}$, because otherwise, at least if $2\sqrt{e}$ is not a limit value for $y_n^2$, then we have $y_{n+1}^2 -y_n^2 \geqslant \frac{C}{n+1}$ and then $y_n$ is unbounded, which is a contradiction. Therefore, this proves at least that $2\sqrt{e}$ is a limit value of $y_n^2$, and it remains to prove that it is the only one. Assuming $y_{\varphi(n)}^2 \to \ell \neq 2\sqrt{e}$ we deduce that $y_{\varphi(n)+1}^2-y_{\varphi(n)}^2 \geqslant \frac{C}{n}$ but from there I am not able to derive any contradiction ...
 A: Edit: here is a simplified argument. It relies on the same idea as the original one, but the technique is much easier. The improvement is to replace the asymptotic identity with $x_n^2H_n=\sum{...}+O(n)$ with a simpler argument based on the growth of $x_n$.
Indeed, with the notations of the original argument (we forget the lemmas but consider ourselves at point (**)), if $n=p_m$ is large enough, then for all but $(1-4\sqrt{\epsilon})n$ of the integers $k$ between $1$ and $n$, $x_k > a'\sqrt{k}$. Thus, there is $n > q_m > n-4n\sqrt{\epsilon}$ with $x_{q_m}/\sqrt{q_m} \geq a'$, then $1 \leq \frac{x_n}{x_{q_m}} \leq \frac{\alpha^-+o(1)}{a'}\left(1-4\sqrt{\epsilon}\right)^{1/2}$, and thus $a' \leq \alpha^-(1-4\sqrt{\epsilon})^{-1/2}$. Take $\epsilon \rightarrow 0$ and we find $\alpha^+ \leq \alpha^-$.

This is probably not the simplest proof, but I think it works.
Let $G_n = \left(\prod_{k=1}^n{x_k}\right)^{1/n}$. We will use many times that $x_n=\Theta(G_n)=\Theta(n^{1/2})$.
Let $\alpha^{\pm}$ be the liminf ($\alpha^-$) and limsup ($\alpha^+$) of $x_n/\sqrt{n}$, we already know $0 < \alpha^- < \alpha^+ < \infty$.
Let, similarly, $\beta^{\pm}$ be the liminf and limsup of $G_n/\sqrt{n}$. By Stirling, we easily see $\alpha^-e^{-1/2} \leq \beta^- \leq \beta^+ \leq \alpha^+e^{-1/2}$.
But let $t < \beta^-$, then for all large enough $n$, $x_{n+1}-x_n < 1/(t\sqrt{n})$, thus for all large enough $n$, $x_n \leq (2/t)\sqrt{n}$. Thus $\alpha^+ \leq 2/\beta^-$, and similarly $\alpha^- \geq 2/\beta^+$.
As $\beta^- \geq \alpha^-e^{-1/2}$ (resp. $\beta^+ \leq \alpha^+e^{-1/2}$), it follows $\alpha^+ \leq 2e^{1/2}/\alpha^-$ (resp. $\alpha^- \geq 2e^{1/2}/\alpha^+$) and therefore $\alpha^+\alpha^-=2e^{1/2}$.
We have moreover (from the above equations) $\alpha^-\beta^+ \geq 2$, hence $\beta^+ \geq 2/\alpha^-=\alpha^+e^{-1/2}$, and thus (with a similar reasoning) we have $\alpha^- = \beta^-e^{1/2} \leq \beta^+e^{1/2}=\alpha^+$ with $\alpha^-\alpha^+=2\sqrt{e}$.

That was the easy part. Now it gets very ugly.
Let $H_n=\sum_{k=1}^n{k^{-1}}$, let $S_n=\sum_{k=1}^n{\frac{1}{x_kG_k}}$.
Lemma: $S_n=\frac{H_n}{2}+O(1)$.
Proof: $(S_n-S_{n-1})-(H_n-H_{n-1})=\frac{2}{x_nG_n}-\frac{1}{n}=2\ln{\frac{x_{n+1}}{x_n}}-\ln{\frac{n+1}{n}}+O(n^{-2})$, which concludes.

Lemma: The following estimate holds, for large enough $n$: $$\frac{x_n^2H_n}{2}=\sum_{k=1}^n{\frac{H_kx_k}{G_k}}+O(n). (*)$$
Proof: $$0.5x_n^2=0.5\sum_{k=1}^{n-1}{x_{k+1}^2-x_k^2}+O(1)=\sum_{k=1}^{n-1}{x_k^2\frac{1}{G_kx_k}}+\sum_{k=1}^{n-1}{G_k^{-2}}+O(1).$$
Thus $$0.5x_n^2 = O(\ln{n})+\sum_{k=1}^n{x_k^2(S_k-S_{k-1})}=O(\ln{n})+\sum_{k=1}^n{x_k^2S_k}-\sum_{k=1}^{n-1}{x_{k+1}^2S_k},$$ thus $0.5x_n^2=O(\ln{n})-\sum_{k=1}^{n-1}{(x_{k+1}^2-x_k^2)S_k}+ x_n^2S_n.$
Therefore $O(n)=x_n^2\frac{H_n}{2}+O(x_n^2)-\sum_{k=1}^{n-1}{\frac{2x_k}{G_k}\frac{H_k}{2}}-\sum_{k=1}^{n-1}{\frac{H_k}{2G_k^2}}+\sum_{k=1}^{n-1}{(S_k-0.5H_k)(x_{k+1}^2-x_k^2)},$ and by the first lemma, our estimates on $G,x$ and the usual $H_n \sim \ln{n}$, we find the result (adding the term in $n$ has no issue).

The liminf on average of $x_n/G_n$ is the liminf of $x_n^2/2n$ so is $(\alpha^-)^2/2$; but its liminf is above $\alpha^-/\beta^+=(\alpha^-)^2/2$, so the liminf of $x_n/G_n$ is $\alpha^-/\beta^+$ and thus there is a sequence $p_n$ with $x_{p_n} \sim \alpha^-\sqrt{p_n}$ and $G_{p_n} \sim \beta^+\sqrt{p_n}$.
Let $y_n$ be the logarithm of $\frac{x_n}{\sqrt{n}}$ and $C_n=\frac{1}{n}\sum_{k=1}^n{y_k}$, so that the limit points of $y_{p_n}$ are between $\gamma^-=\ln{\alpha^-}$ and $\gamma^+=\ln{\alpha^+}$. By Stirling, and the above, $C_n = \ln{\frac{e^{1/2}G_n}{\sqrt{n}}}+o(1)$, so that $y_{p_n} \rightarrow \gamma^-$, $C_{p_n} \rightarrow \gamma^+$.
Let $0.1 > \epsilon > 0$, we claim that for large enough $n$, there are at most $3\sqrt{\epsilon}p_n$ integers $1 \leq k \leq p_n$ such that $y_k \leq \gamma^+-\sqrt{\epsilon}$.
Indeed, let $N \geq 1$ be such that $y_n \leq \gamma^++\epsilon$ for $n \geq N$. For $n$ large enough, $\sqrt{\epsilon}p_n > N$ and $\sum_{k=N+1}^{p_n}{(\gamma^++\epsilon-y_k)} \leq 2(p_n-N)\epsilon$. By Markov inequality, there must be at most $2(p_n-N)\epsilon/(\sqrt{\epsilon}+\epsilon) \leq 2p_n\sqrt{\epsilon}$ integers $k$ with $y_k \leq \gamma^+-\sqrt{\epsilon}$.
But now, consider the equation (*) with $n=p_m$ for a large enough $m$. There is a subset $S$ made with all the integers from $1$ to $n$ except at most $3\sqrt{\epsilon}n$ such that $x_k \geq a'\sqrt{k}$ for $k \in S$, $a'=\alpha^+e^{-\epsilon}$. (**)
Let us take $b' > \beta^+$, then for finitely many $k$, $G_k < b'\sqrt{k}$. So for $n=p_m$ large enough, there is a set $S$ of at least $n-4\sqrt{\epsilon}n$ integers between $1$ and $n$ such that, for each $k \in S$, $G_k < b'\sqrt{k}$, $x_k > a'\sqrt{k}$.
Note that $\sum_{k \in S}{H_k} \geq (1+o(1))\sum_{1 \leq k \leq (1-4\sqrt{\epsilon})n} {\ln{k}} \geq (1+o(1))n(1-4\sqrt{\epsilon})\ln{n}$.
When considering (*), it follows that for large enough $n=p_m$, $(\alpha^-)^2/2 n\ln{n} \sim x_n^2H_n/2 \geq (1+o(1))a'/b'(1-4\sqrt{\epsilon})n\ln{n}+O(n)$. Therefore, $(\alpha^-)^2/2 \geq a'/b'(1-4\sqrt{\epsilon})$.
Now take $\epsilon \rightarrow 0$, $b' \rightarrow \beta^+$, and thus $(\alpha^-)^2/2 \geq \alpha^+/\beta^+=\sqrt{e}$, thus $(\alpha^-)^2 \geq 2\sqrt{e} = \alpha^-\alpha^+$ and thus $\alpha^-=\alpha^+$.
