4
$\begingroup$

Let $(a_n)_{n\in\mathbb{N}}$ be a positive sequence. Assume, that

  1. $\lim_{n \to \infty} a_n = \infty$.
  2. $\forall\ \zeta > 0, \exists\ n=n(\zeta) \geq 1: a_n < \zeta.$

From 1. we know, that $\forall\ C > 0, \exists\ N = N(C) \geq 1: a_n \geq C, \forall\ n \geq N.$

The difference $N(C) - n(\zeta)$ can be interpreted as the 'recovery time', i.e. the time the sequence needs from being arbitrarily small to become larger than an arbitrary positive constant $C$ for the rest of its existence.

Question: Can somebody think of an example where $\forall\ C > 0: N(C) - n(\zeta)$ is arbitrarily large? (Is this even possible?)

$\endgroup$
2
  • 9
    $\begingroup$ 1. says $\liminf_n a_n = \infty$ and 2. says $\liminf_n a_n = 0$. $\endgroup$
    – copper.hat
    Jun 10, 2018 at 0:08
  • 1
    $\begingroup$ You could rephrase (1) to be in terms of lim sup, but it would alter what you're trying to find. But I think you could find a sequence that spends an arbitrarily long time near zero before becoming larger than previously. Something like $a_k=1/k$ if $k\ne n!$ and $a_{n!}=n$ otherwise, for example. $\endgroup$
    – Teepeemm
    Jun 10, 2018 at 1:51

2 Answers 2

22
$\begingroup$

There is no sequence satisfying both of your conditions 1 and 2. Proof: given that $a_n \to \infty$, there is a constant $N = N(1)$ such that $a_n \geq 1$ for all $n \geq N$. Then there are only finitely many terms smaller than 1, and you can choose $\zeta$ to be smaller than any of these.

$\endgroup$
0
5
$\begingroup$

I don't believe there exists a sequence satisfying both 1. and 2.

The sequence $n(1), n(1/2), n(1/3), \ldots$ (using the notation in condition 2.) is a sequence of positive integers, and thus has a monotone subsequence. This sequence cannot be eventually constant (i.e. $n(1/k)=n_*$ for all sufficiently large $k$) else $a_{n_*} < 1/k$ for all large $k$ which contradicts the fact that $a_{n_*} > 0$. Thus $n(1), n(1/2), n(1/3), \ldots$ has an increasing subsequence, so there is a subsequence of $(a_n)_n$ that decreases to zero. This contradicts 1.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .