Consider a sequence $(x_n)_{n\geq 0}$ in $(0,1)$ such that $4x_n(1-x_{n+1})>1$ for all $n$. Show that the sequence is monotone and find its limit.

What I attempted:-
I started in an informal way.
The sequence is bounded in $(0,1)$. We are given that
$$4x_n(1-x_{n+1})>1\Rightarrow x_n(1-x_{n+1}) > \frac{1}{4}\Rightarrow x_n(1-x_{n+1}) >0\\ \Rightarrow x_n>x_n x_{n+1} \Rightarrow x_n\cdot 1>x_n x_{n+1} \Rightarrow x_n>x_{n+1} \quad\text{(since $x_n<1$)}.$$
So, the sequence is monotonic decreasing, which eventually implies that the sequence converges. Let $p$ be its limit.

Then for large $n$,
$$4p(1-p)>1\Rightarrow (2p-1)^2 <0\Rightarrow p<\frac{1}{2}.$$
Am I in the right track? I urge the readers to mark it as duplicate if it has already been raised.

  • 1
    $\begingroup$ You had to write :(Since $x_{n+1}<1$ not since $x_n<1$. $\endgroup$ – hamam_Abdallah Jun 25 '18 at 14:18

If $x_n\to p$ then it should be $$4x_n(1-x_{n+1})>1\implies 4p(1-p)\geq 1\implies (2p-1)^2 \leq 0\implies p=\frac{1}{2}.$$ Check again your proof of the monotone property. Note that if $x_n>0$ then $$x_{n+1}<1-\frac{1}{4x_n}=x_n-x_n+1-\frac{1}{4x_n}\leq x_n-\frac{(2x_n-1)^2}{4x_n}\leq x_n.$$

  • $\begingroup$ Does it means that for large $n$, this inequality may become an equality? Is their no possibility that the proposed limit is less than $\frac{1}{2}$( say $\frac{1}{3}$) ? $\endgroup$ – user440191 Jun 25 '18 at 14:05
  • $\begingroup$ Note that $1/n>0$ but $1/n\to 0$. $\endgroup$ – Robert Z Jun 25 '18 at 14:08
  • $\begingroup$ In order to satisfy the given inequality the whole sequence has to be $>1/2$. $\endgroup$ – Robert Z Jun 25 '18 at 14:31

If $(x_n)$ satisfies the condition, then $(1-x_n)$ satisfies also the condition.

we should have $$\lim_{n\to\infty}x_n=\lim_{n\to\infty}(1-x_n)$$ or $$\lim_{n\to+\infty} x_n=\frac12$$


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