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

Problem. Let $$(x_n)$$ be the sequence defined by $$x_0=\frac{1}{2}$$ and $$x_{n+1}=x_n^3-x_n^2+1$$ for any $$n\in \mathbb{N}\cup \{0\}$$. Find

$$\lim_{n\to \infty} (x_0 x_1 \cdots x_n)\sqrt{n}.$$

According to the answer sheet, this limit equals $$1$$. However, I can't manage to solve it. Here is what I've done.

Obviously, $$x_{n+1}-x_n=(x_n-1)^2(x_n+1)>0$$ (it is easy to observe that all the terms of the sequence are positive), so $$(x_n)$$ is a strictly increasing sequence.

Let us now prove by induction on $$n$$ that $$x_n<1$$ for all $$n\in \mathbb{N}\cup \{0\}$$. The base case is obvious, so suppose it holds for $$n$$ and prove it for $$n+1$$. $$x_{n+1}=x_n^2(x_n-1)+1<1$$ by the induction hypothesis and we are done.

Hence, $$(x_n)$$ is monotone and bounded, so it is convergent. It is easy to see now that $$\displaystyle \lim_{n\to \infty}x_n=1$$.

Now I pretty much got stuck. I tried to use the epsilon definition of a limit, trying to exploit $$\displaystyle \lim_{n\to \infty}x_n=1$$, but it didn't help. Maybe I should use Stolz–Cesàro on the limit that I want to compute?

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

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 $$(3)$$ $$1>y_n\geq \sqrt{\frac{2n}{2n+2\log{(n-1)}+C}}\to1, n\to\infty$$

This question is similar to previous MSE questions including question 2675217, question 2776443, question 2861768.

Define the recursion $$a_{n+1}=f(a_n)\;\;\text{ where }\;\; f(x):=x(1-x)^2. \tag{1}$$ Define the sequence $$\, x_n := 1-a_n.\,$$ Verify that $$\, x_{n+1}=x_n^3-x_n^2+1.\,$$

Assume an asymptotic expansion $$a_n = \sum_{k=0}^\infty p_k(c,z)y^{k+1}, \;\; y := 1/n,\quad z := \log(y) \tag{2}$$ where $$\,p_k(c,z)\,$$ is a polynomial of degree $$\,k\,$$ in $$\,c\,$$ and $$\,z.\,$$

Substitute equation $$(2)$$ into equation $$(1)$$ and solve for the polynomials $$\,p_k\,$$ to get $$p_0\!=\!\frac12,\, p_1\!=\!\frac{c\!+\!3z}8,\, p_2\!=\!\frac{(5\!+\!3c\!+\!c^2)\!+\! (9\!+\!6c)z\!+\!9z^2}{32}. \tag{3}$$

Define the sequence $$y_n:=(x_0x_1...x_n)\sqrt{n}. \tag{4}$$ Use equation $$(1)$$ and the definition of $$\,x_n\,$$ to verify that $$y_n=\sqrt{na_{n+1}/a_0}. \tag{5}$$

Use equation $$(2)$$ to get $$\lim_{n\to\infty} y_n = \sqrt{\frac{p_0}{a_0}} = \frac1{\sqrt{2a_0}}. \tag{6}$$ When $$\,x_0=a_0=\frac12,\,$$ the limit is $$1$$. Also, $$\,c\approx -8.205896246.$$