I am stuck with this problem that appeared in the Undergrad Brazilian Math Olympiad from 2017. The problem is:
let $x_n$ be a strictly positive sequence that $x_n\rightarrow 0$.
Suppose that exists $c>0$ that $|x_{n+1}-x_n|\leq c x_n ^2$ for all
$n\in\mathbb{N}$. Show that there is $d>0$ that $n x_n\geq d$ for all
$n\in \mathbb{N}$.
I tried using the Stolz-Cesàro lemma, but didn't help me very much. Does anyone have a hint? Thanks!
EDIT:
Let me give some context of my idea. For the Stolz-Cesàro lemma the given sequence $x_n$ needs to be strictly decreasing, since it $x_n\rightarrow 0$ and $x_n>0$. Well, I don't know if that is true, the best thing I've got was: given $\varepsilon>0$ it is true that $(1-\varepsilon)a_n<a_{n+1}<(1+\varepsilon)a_n$ for sufficiently large $n$. One could help me on that.
Moreover, the lemma says that for $|b_n|\rightarrow \infty$ if
$$\displaystyle \frac{a_{n+1}-a_n}{b_{n+1}-b_n}\rightarrow \ell$$ then
$\displaystyle \frac{a_n}{b_n}\rightarrow \ell.$
Supposing that $x_n$ is strictly decreasing, than I can choose $a_n=n$ and $b_n=1/x_n$. That way I would have
$$c_n=\frac{(n+1)-n}{\frac{1}{x_{n+1}}-\frac{1}{x_{n}}}=\frac{1}{\frac{1}{x_{n+1}}-\frac{1}{x_{n}}}.$$
If it is possible to show that this sequence $c_n$ converges to some positive number I would have the result.
But with these assumptions (including that $x_n$ is strictly decreasing) the best thing that I've got was:
$$\frac{1}{c}(1-x_n)\leq \frac{1}{\frac{1}{x_{n+1}}-\frac{1}{x_{n}}}=\frac{x_{n+1}x_n}{x_n-x_{n+1}}.$$
At this point there are two things that I don't know: (1) $x_n$ strictly decreases and (2) how do I find a comparison (if there is any) to show that $\displaystyle \frac{x_{n+1}x_n}{x_n-x_{n+1}}< d_n$, where $d_n\rightarrow 1/c$.
One last thing that I noticed is that the hypothesis $|x_{n+1}-x_n|\leq c x_n ^2$ implies that $x_{n+1}/x_n\rightarrow 1$ and
$$f_n=\frac{|x_{n+1}-x_n|}{x_n ^2}$$
has a convergent subsequence. These facts implies that
$$\frac{x_{n+1}x_n}{x_n-x_{n+1}}$$ also has a convergent subsequence.
That's it, please help!
Best Answer
We're to show that there exists such $d>0$ that $\frac{1}{nx_n}<\frac{1}{d}$ i.e. that $\frac{1}{nx_n}$ is bounded from above.
Consider $y_n=\frac{1}{cnx_n}$ i.e. $x_n=\frac{1}{cny_n}$ then the inequality $|x_{n+1}-x_n|\le cx_n^2$ becomes $$\left|\frac{1}{c(n+1)y_{n+1}}-\frac{1}{cny_n}\right|\le c\frac{1}{c^2n^2y_n^2}$$ and we can cancel $c$. After some rearrangements the inequality becomes $$\frac{1}{n y_n - 1} + 1 \ge (n+1)y_{n+1}-ny_n\ge \frac{1}{1 + n y_n} - 1$$ then, noting $ny_n\to +\infty$ as $ny_n=\frac{1}{cx_n}$ and $x_n\to +0$, we have $\frac{1}{ny_n-1}+1\to 1$ thus LHS is bounded by some constant $C$ from above and we can write $$C\ge \frac{1}{n y_n - 1} + 1 \ge (n+1)y_{n+1}-ny_n$$ $$C\ge (n+1)y_{n+1}-ny_n$$ summing up for $n=1,\ldots,\,m$ we have $$Cm\ge (m+1)y_{m+1}-y_1$$ $$C(m+1)\ge Cm\ge (m+1)y_{m+1}-y_1$$ $$C\ge y_{m+1}-\frac{y_1}{m+1}$$ $$y_1+C\ge\frac{y_1}{m+1}+C\ge y_{m+1}$$ i.e. $y_{m+1}$ is bounded from above. QED.
"Some rearrangements":
$$\left|\frac{1}{c(n+1)y_{n+1}}-\frac{1}{cny_n}\right|\le c\frac{1}{c^2n^2y_n^2}$$ $$-\frac{1}{n^2y_n^2}\le \frac{1}{(n+1)y_{n+1}}-\frac{1}{ny_n}\le \frac{1}{n^2y_n^2}$$ $$\frac{1}{ny_n}-\frac{1}{n^2y_n^2}\le \frac{1}{(n+1)y_{n+1}}\le \frac{1}{ny_n}+\frac{1}{n^2y_n^2}$$ $$\frac{ny_n-1}{n^2y_n^2}\le \frac{1}{(n+1)y_{n+1}}\le \frac{ny_n+1}{n^2y_n^2}$$ Now we consider only that $y_n$ for which $ny_n-1>0$, the other are already bounded from above by $\frac 1n$. $$\frac{n^2y_n^2}{ny_n-1}\ge (n+1)y_{n+1}\ge \frac{n^2y_n^2}{ny_n+1}$$ $$\frac{n^2y_n^2-ny_n(ny_n-1)}{ny_n-1}\ge (n+1)y_{n+1}-ny_n\ge \frac{n^2y_n^2-ny_n(ny_n+1)}{ny_n+1}$$ $$\frac{ny_n}{ny_n-1}\ge (n+1)y_{n+1}-ny_n\ge \frac{-ny_n}{ny_n+1}.$$