Finding $\displaystyle \lim_{n\to \infty} (x_0 x_1 \cdots x_n)\sqrt{n}$ where $x_{n+1}=x_n^3-x_n^2+1$, $x_0=\frac{1}{2}$

Let's note as $$y_n=(x_0x_1...x_n)\sqrt{n} \tag{1}$$ From $$\frac{x_{n+1}-1}{x_{n}-1}=x_n^2$$ we have $$\frac{x_{n+1}-1}{x_{n}-1}\cdot\frac{x_{n}-1}{x_{n-1}-1}=\frac{x_{n+1}-1}{x_{n-1}-1}=x_n^2 x_{n-1}^2$$ then $$\frac{x_{n+1}-1}{\frac{1}{2}-1}=x_n^2 x_{n-1}^2...x_0^2 \iff x_n^2 x_{n-1}^2...x_0^2=2(1-x_{n+1})\iff$$ $$y_n=\sqrt{2n(1-x_{n+1})} \tag{2}$$

Proposition 1. For $n\geq1$ $$x_n \geq \frac{2n-1}{2n}$$

To begin with, $f(x)=x^3-x^2+1$ is ascending on $\left[\frac{2}{3},1\right]$, easy to check by taking the derivative.

By induction

• $x_1=\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^2+1=\frac{7}{8}>\frac{2\cdot1-1}{2\cdot1}=\frac{1}{2}$
• $x_2=\left(\frac{7}{8}\right)^3-\left(\frac{7}{8}\right)^2+1=\frac{463}{512}>\frac{2\cdot2-1}{2\cdot2}=\frac{3}{4}>\frac{2}{3}$
• now let's assume $x_n>\frac{2n-1}{2n}>\frac{2}{3}$, then ($f(x)$ is ascending!) $$x_{n+1}=f(x_n)\geq f\left(\frac{2n-1}{2n}\right)=\\ \frac{8n^3-4n^2+4n-1}{8n^3}\geq \frac{2n+1}{2n+2}$$ since $$\frac{8n^3-4n^2+4n-1}{8n^3}-\frac{2n+1}{2n+2}=\frac{3n-1}{8n^3(n + 1)}>0$$ not too difficult to check.

And we are done. As a result, from $(2)$ $$0<y_n\leq \sqrt{\frac{2n}{2n+2}}<1\tag{3}$$

Proposition 2. For $n\geq1$ $$y_{n+1} > y_n$$

From $(1)$ $$\frac{y_{n+1}}{y_n}=x_{n+1}\sqrt{\frac{n+1}{n}} \tag{4}$$ $$\overset{\text{Pr1}}{\geq} \frac{2n+1}{2n+2}\cdot \sqrt{\frac{n+1}{n}}= \frac{2n+1}{2\sqrt{n(n+1)}}>1$$

The sequence $(y_n)_{n>0}$ is bounded and ascending, so it has a limit. Is it $1$ though? We will notice that

$$\frac{1}{1-x_{n+1}}=\frac{1}{x_n^2 (1-x_n)}=\frac{1}{x_n^2}+\frac{1}{x_n}+\frac{1}{1-x_n}$$ then $$\frac{1}{1-x_{n+1}}-\frac{1}{1-x_n}=\frac{1}{x_n^2}+\frac{1}{x_n}\leq ...$$ using Proposition 1 $$...\leq \left(\frac{2n}{2n-1}\right)^2+\frac{2n}{2n-1}= \left(1+\frac{1}{2n-1}\right)^2+1+\frac{1}{2n-1}=\\ 2+\frac{3}{2n-1}+\frac{1}{(2n-1)^2}\leq 2+\frac{4}{2n-1}\leq 2+\frac{2}{n-1}$$ and $$\frac{1}{1-x_{n+1}}-\frac{1}{1-x_2}= \sum_{k=2}^{n}\left(\frac{1}{1-x_{k+1}}-\frac{1}{1-x_k}\right)\leq\\ 2(n-2)+2\sum_{k=2}^n\frac{1}{k-1}=2(n-2)+2\sum_{k=1}^{n-1}\frac{1}{k}$$ Basically (where $C$-const) $$\frac{1}{1-x_{n+1}}\leq 2n+2\log{(n-1)}+C$$ and using $(2)$ and $(3)$ $$1>y_n\geq \sqrt{\frac{2n}{2n+2\log{(n-1)}+C}}\to1, n\to\infty$$