Let $(u_n)_{n\in \mathbb N^*}$ a non-zero natural number sequence, with : $$\forall n \in \mathbb N^*, |u_{n+1}-u_n|\leq \frac{u_{n}u_{n+1}}{(n+1)^2}$$

Is it true that : $u_{n}$ is bounded, if and only if, $\exists n\in \mathbb N^*,u_n=n$ ?

Source : les dattes à Dattier

  • $\begingroup$ what does $\mathbb{N}^*$ stand for $\endgroup$ – user394255 Jan 24 '17 at 17:34
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    $\begingroup$ the non-zero natural number set. $\endgroup$ – Dattier Jan 24 '17 at 17:54
  • $\begingroup$ The sequence defined by $u_n=n$ does not fit your condition. Neither is it bounded. $\endgroup$ – Ivan Neretin Jan 25 '17 at 18:09
  • $\begingroup$ $|n+1-n|=1$ and $\frac{n(n+1)}{(n+1)^2}<1$ $\endgroup$ – Dattier Jan 25 '17 at 18:17
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    $\begingroup$ WTF: Why the spree editions naming some results after yourself? A system flag was raised because to the site SW this was vandalism. If it was meant to be a joke, it is IMHO in a bit poor taste :-/ $\endgroup$ – Jyrki Lahtonen Apr 5 '17 at 20:25

Let $(u_n : n\geq 1)$ be a sequence of positive integers such that

$$ \forall n \geq 1, \qquad |u_{n+1} - u_n| \leq \frac{u_n u_{n+1}}{(n+1)^2}. \tag{*}$$

Notice that this condition implies

$$ \forall n \geq 1, \qquad \left| \frac{1}{u_n} - \frac{1}{u_{n+1}} \right| < \frac{1}{n} - \frac{1}{n+1}. \tag{$\diamond$}$$

It turns out that $(\diamond)$ is the key property.

Lemma. $(u_n)$ is unbounded if and only if $u_n > n$ for all $n$.

Before proving this lemma, let us check how this implies OP's claim. If $u_n = n$ for some $n$, then $(u_n)$ is bounded by the lemma above.

Conversely, let $(u_n)$ be bounded. By the lemma, we can find the smallest $n$ for which $u_n \leq n$ holds. We claim that $u_n = n$.

If $n = 1$, then we are done since $1 \leq u_n \leq n = 1$. If $n > 1$, then we have $u_{n-1} \geq n \geq u_n$. If it happens that $u_n < n$, then $u_n \leq n-1$. Thus by $(\diamond)$,

$$ \frac{1}{n-1} - \frac{1}{u_{n-1}} \leq \frac{1}{u_n} - \frac{1}{u_{n-1}} < \frac{1}{n-1} - \frac{1}{n}. $$

This implies that $u_{n-1} < n$, which contradicts $u_{n-1} \geq n$. Therefore $u_n = n$.

Proof of Lemma. One direction is trivial, so we assume that $(u_n)$ is unbounded and prove that $u_n > n$. By the assumption, there exists a subsequence $(u_{n_j})$ such that $u_{n_j} \to \infty$. Then by $(\diamond)$,

$$ \frac{1}{u_n} = \lim_{j\to\infty} \left| \frac{1}{u_n} - \frac{1}{u_{n_j}} \right| \leq \sum_{k=n}^{\infty}\left| \frac{1}{u_k} - \frac{1}{u_{k+1}} \right| < \sum_{k=n}^{\infty} \left( \frac{1}{k} - \frac{1}{n+1}\right) = \frac{1}{n} $$

and hence $u_n > n$.


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