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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!

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  • $\begingroup$ Is this the olympiad page? Don't they publish solutions? Thanks. $\endgroup$ – Alexey Burdin Jul 20 at 23:22
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    $\begingroup$ I don't understand the vote to close this question. $\endgroup$ – Robert Shore Jul 20 at 23:27
  • $\begingroup$ They do publish some of the problems solutions, but this one isn't published. $\endgroup$ – Vinnie Carvalho Jul 21 at 0:19
  • $\begingroup$ @RobertShore I'm guessing because of "shows no effort". IE It is "missing context and details". $\endgroup$ – Calvin Lin Jul 21 at 1:37
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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 $\frac 1n$. $$\frac{n^2y_n^2}{ny_n-1}\ge (n+1)y_{n+1}\ge \frac{n^2y_n^2}{ny_n+1}$$ $$\frac{n^2y_n^2-ny_n(ny_n-1)}{ny_n-1}\ge (n+1)y_{n+1}-ny_n\ge \frac{n^2y_n^2-ny_n(ny_n+1)}{ny_n+1}$$ $$\frac{ny_n}{ny_n-1}\ge (n+1)y_{n+1}-ny_n\ge \frac{-ny_n}{ny_n+1}.$$

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  • $\begingroup$ Thank you for your help! $\endgroup$ – Vinnie Carvalho Jul 21 at 2:48
  • $\begingroup$ You're welcome.) Spent 3.5 hours so not-a-contest solution) Contests are 4h-6h likely. $\endgroup$ – Alexey Burdin Jul 21 at 2:51
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[This seems a lot easier than I expected, so there might be errors in it. If so, please let em know where.]

These steps can be demystified by referencing the subsequent block of text

  1. Pick $N$ such that $ \forall n > N$, $x_n < \frac{ 1}{2c}$.
  2. Set $ k = \min ( \frac{1}{2c}, Nx_N ) $. Observe that $\frac{N}{N+1} \geq \frac{1}{2} \geq ck$ and $ Nx_N \geq k$.
  3. Hence $(N+1) x_{N+1} \geq (N+1)( x_N - cx_N^2) \geq (N+1)(\frac{k}{N} - \frac{ ck^2}{N^2}) = k + \frac{k( \frac{N}{N+1} - ck ) }{N^2(N+1)} \geq k$.
  4. Also, $ \frac{N+1}{N+1+1} \geq \frac{1}{2} \geq ck$.
  5. Proceed by induction to conclude that $ n x_n \geq k$.

We claim that under suitable conditions (to be determined), if $ n x_n \geq k$, then $(n+1) x_{n+1} \geq k $. If so, the result follows by induction.

Which conditions make sense?

  1. We have $x_{n+1} \in ( x_n - c x_n^2, x_n + cx_n^2) $.
  2. We have $ \frac{k}{n} < x_n$.
  3. We will likely want $x - c x^2$ to be increasing, which requires $ x_n < \frac{1}{2c}$. This can be satisfied as $ \lim x_n = 0 $.
  4. Henceforth, we assume $ \frac{k}{n} < x_n < \frac{1}{2c}$. This necessitates $ 2ck < n$, which can be achieved.
  5. Now, $ (n+1) x_{n+1} > (n+1) \left[ x_n - c x_n^2\right] > (n+1) \left[\frac{k}{n} - \frac{ ck^2 } { n^2 } \right] $. Verify that $ (n+1) \left[\frac{k}{n} - \frac{ ck^2 } { n^2 } \right] \geq k \Leftrightarrow \frac{n}{n+1} \geq ck $.

This gives us all of the conditions that we need.

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  • $\begingroup$ Thank you for your help! $\endgroup$ – Vinnie Carvalho Jul 21 at 2:48
  • $\begingroup$ The parts order from this revision made more sense for me. $\endgroup$ – Alexey Burdin Jul 21 at 2:49
  • $\begingroup$ @AlexeyBurdin I agree with you. It's the difference between presenting a clean direct solution for the olympiad vs motivating the approach / values chosen. I just started writing stuff down and it follows immediately (which is why I'm questioning if there is an error). $\endgroup$ – Calvin Lin Jul 21 at 2:54

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