Sequence of positive terms with $\lim a_n=0$ and $ c>0$ such that $|a_{n+1}-a_n|\leq c\cdot a_n^2$.

Question

I am dealing with the test of the OBM (Brasilian Math Olimpyad), University level, 2017, phase 2.

As I've said at others topic (questions 1, 2 and 3), I hope someone can help me to discuss this test.

The question 4 says:

Let be $(a_n)_{n\geq 1}$ a sequence of positive terms with $\lim a_n=0$ such that, for a certain $c>0$ and for all $ n\geq 1$, $|a_{n+1}-a_n|\leq c\cdot a_n^2$. Prove that exists $d>0$ with $n\cdot a_n\geq d, \forall n\geq 1$.

Well, I unfortunately couldn't do a lot in this question. I've thinked about Cauchy, I've done some passages with the inequalities, but I couldn't get anything substantial. Thanks for the help.

score 2 · Accepted Answer · answered Sep 16 '18 at 22:23

2

Since $\lim_{n \to +\infty} a_n = 0$ we have that there exist $n_0$ such that for any $n \geq n_0$ we have $a_n > ca_{n}^2$ and $1 > 2ca_n$.

Now for $n$ big enough $$a_{n+1} \geq a_n - c \cdot a_n^2 > 0$$ Consider $d>0$ such that the inequality is true until $n_0$ and $\frac{d}{n_0} \cdot 2c < 1$ and $cd < \frac{1}{3}$ (it clearly exists since there are only finitely many values of $na_n$ for $n \leq n_0$).

Suppose now $a_{n+1} < \frac{d}{n+1}$ and $a_n \geq \frac{d}{n}$ for some $n \geq n_0$.

Then, since we have $n \geq n_0$ and so the function $f(x) = x-cx^2$ is increasing in the interval we are considering, $$ \frac{d}{n+1} > \frac{d}{n} - \frac{cd^2}{n^2}$$ And so $$ dn^2 > dn(n+1) - cd^2(n+1) \iff cd(n+1) > n$$ Thus $$ 3n < n+1 \iff n < \frac{1}{2}$$ Which is absurd.

Therefore $na_n \geq d$ for any $n \in \mathbb{N}$

answered Sep 16 '18 at 22:23

LucaMac

3,355

I finally understand! Your answer was clear and very different that I've supposed, thanks very much. Soon I'll post more contest questions that I'm dealing. – Quiet_waters Sep 17 '18 at 17:10
If I ask you about how the ideia for this question appear, you would have some clue, or did the ideia just "appear" in your mind (obviously, very trained)? I ask because I was really stucked at this question. – Quiet_waters Sep 17 '18 at 17:16
1

I just thought "we have to prove $a_n>$ something, and I want to use induction so let's write the inequality as $a_{n+1} >$ something depending on $a_n$" – LucaMac Sep 17 '18 at 21:34
Right, thanks for the comment! – Quiet_waters Sep 17 '18 at 22:41
Why $n<1/2$ is absurd? – João Víctor Melo Jan 14 '22 at 13:58
When did you use the fact that $|a_{n+1}-a_n| \leq c(a_n)^2$? – João Víctor Melo Jan 14 '22 at 14:03

Sequence of positive terms with $\lim a_n=0$ and $ c>0$ such that $|a_{n+1}-a_n|\leq c\cdot a_n^2$.

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