# A limit of a recursive sequence

I was wondering if someone can give me a hint on the following problem:

Given $0<x_1<1$ and $x_{n+1}=x_n(1-x_n)$ for $n\ge 1$, prove that $$\lim_{n\to\infty}nx_n=1.$$

So far, I am able to confirm that this limit exists (indeed, the sequence $\{nx_n\}$ is increasing and bounded above). But it seems that this limit cannot be obtained by using the recursive formula in a simple way (like transforming and taking limit on both sides, which always leads to an identity). Any hint will be highly appreciated.

• This question is a weaker form of the MSE question 2861888 which asks for the limit of $n(1-nx_n)/\log n$ which first requires that limit of $n x_n$ is 1. – Somos Aug 5 '18 at 12:23

## 2 Answers

According to your preliminary work the sequence $\{nx_n\}$ is positive, increasing and bounded above, therefore we may assume that $\lim_{n\to\infty}nx_n=L\in\mathbb{R}^+$.

Then by Stolz-Cesaro Theorem (see also Stolz-Cesaro Theorem, 0/0 Case) $$L=\lim_{n\to \infty}nx_n=\lim_{n\to \infty}\frac{x_n}{1/n}=\lim_{n\to \infty}\frac{x_{n+1}-x_n}{\frac{1}{n+1}-\frac{1}{n}}=\lim_{n\to \infty}\frac{-x_n^2}{-\frac{1}{(n+1)n}}=L^2.$$ Hence $L=L^2$ and we may conclude that $L=1$.

Proof. $\blacktriangleleft$ First we show that $x_n \searrow 0$. Since $x_1 \in (0, 1)$, we use AM-GM inequalityvto deduce that $$x_2 = x_1(1-x_1) \leqslant \frac 14.$$ By induction, we can show that $x_n \in (0,1)$ for all $n$. Hence $$x_{n+1}= x_n (1-x_n) \leqslant x_n.$$ Therefore $0 \leqslant x_n \searrow$, then $x_n$ converges. Setting the limit as $A$, we have $$A = A (1-A),$$ hence $A=0$.

Now transform the recursive relation: take reciprocal, then $$\frac 1 {x_{n+1} }= \frac 1 {x_n(1-x_n)} = \frac 1{x_n} + \frac 1 {1-x_n},$$ hence $$\frac 1 {x_{n+1}} - \frac 1{x_n} = \frac 1{1-x_n}.$$ Now by Stolz formula, \begin{align*} \lim_n \,nx_n &= \lim \frac n {1/x_n} \\ &= \lim \frac {n+1 - n} {x_{n+1}^{-1} - x_n ^{-1}} \quad [\text{O. Stolz}]\\ &= \lim (1 - x_n) = 1. \blacktriangleright \end{align*}