I think $x_{0}$ $\in$ $(0,1)$

khiemnguyen0620 wrote: I think $x_{0}$ $\in$ $(0,1)$ Oh no : it converges for example for $x_0=10$ while it is $\sim n-1$ (so unbounded) for $x_0=\frac 34$

In fact, answer is likely $x_0\in\left(0,\frac{-1+\sqrt 5}2\right]\cup[1,+\infty)$ (easy to show that this interval gives bounded sequence, but I did not prove clearly that the other cases all are unbounded)

pco wrote: In fact, answer is likely $x_0\in\left(0,\frac{-1+\sqrt 5}2\right]\cup[1,+\infty)$ (easy to show that this interval gives bounded sequence, but I did not prove clearly that the other cases all are unbounded) How did you show that the sequence is bounded in this case?

bnumbertheory wrote: How did you show that the sequence is bounded in this case? In this case, $x_2=\frac{2x_0}{x_0^3+1}\le 1$ and then $x_{n+1}\le x_n\le 1$ $\forall n\ge 2$

Solved with Euler365. We give a series of claims: Claim 1: If $x_n \leq 1$ for $n \geq 1$, then $x_{n+1} \leq 1$. Proof: $x_{n+1} \leq 1$ is equivalent to $(1-x_n)(n^2+n^2x_n-x_n^2) \geq 0$, which is true. $\blacksquare$ Claim 2: If $x_n \leq 1$ for some $n \geq 1$, then either $x_1 \leq 1$ or $x_2 \leq 1$. Proof: Assume FTSOC that the smallest positive integer $n$ such that $x_n>1$ and $x_{n+1} \leq 1$ is at least $2$. Then using $x_n>1$ in th expression for $x_{n+1} \leq 1$, we get $x_n \geq \frac{n^2+n\sqrt{n^2+4}}{2} > n^2$. Now, $x_{n-1}>1$ in the expression of $x_n$ gives $$x_n=\frac{(n-1)^2+1}{x_{n-1}+\frac{(n-1)^2}{x_{n-1}^2}} < (n-1)^2+1$$which is absurd. $\blacksquare$ Solving for $x_1 \leq 1$ and $x_2 \leq 1$ gives $$\boxed{x_0\in\left(0,\frac{-1+\sqrt 5}2\right]\cup[1,+\infty)}$$and clearly these values work. Now we show that no other value of $x_0$ works. Claim 3: If $x_n>1$ for all sufficiently large $n$ and $\{x_n\}$ is bounded, then $\{x_n\}$ is eventually strictly increasing. Proof: Write $$x_{n+1}-x_n=\frac{x_n(x_n-1)(n^2-x_n-x_n^2)}{x_n^3+n^2}$$Since $x_n>1$ eventually, and $\{x_n\}$ bounded implies $x_n^2+x_n<n$ for all sufficiently large $n$, $\{x_n\}$ is eventually strictly increasing. $\blacksquare$ Now assume we have a sequence $\{x_n\}$ which is bounded but is eventually $>1$ always. Then it is eventually strictly increasing, plus bounded so it has a limit $L>1$. But, rewrite $$x_{n+1}=\frac{\left(1+\frac{1}{n^2}\right)x_n^2}{1+\frac{x_n^3}{n^2}}$$Taking $n \rightarrow \infty$ in the above, we get $L=L^2$, which is impossible since $L>1$. Therefore every such bounded sequence must stay below $1$ eventually, and Claim 2 gives us the required values of $x_0$ that work. $\blacksquare$