Eventually self-outgrowing = eventually monotonic?

**jens** · May 31st 2012, 05:38 AM

Hi,

Consider a bounded sequence $(a_n)_{n \in \mathbb{N}}$ whose terms are eventually "outgrown" by other terms of the sequence, i.e., for any index $n$ there exists an $n' > n$ such that $a_{n'} > a_n$ . Does this property imply that $a_n$ is eventually monotonic, i.e., that there exists an index $n_0$ such that $a_n$ is monotonically increasing from index $n_0$ onwards? I can't think of any counterexample. If true, do you know if this a known theorem which I can reference from some standard textbook?

NB: the boundedness assumption is essential, otherwise you can construct simple counterexamples like $a_n = n + (-1)^n$

Thanks,
jens

**girdav** · May 31st 2012, 05:46 AM

$a_{2n}:=1-\frac 1{2n}$ and $a_{2n+1}=0$ .

**jens** · May 31st 2012, 02:49 PM

Thanks for your counterexample!
Sorry for the mess, but I had only sequences in mind whose variations become "tiny", that is, whose $\lim\sup$ and $\lim\inf$ coincide. But I forgot to tell. So let me correct my question:
Assume $a_n$ converges to a finite limit and for any index $n$ , there exists $n' > n$ such that $a_{n'} > a_n$ . Does this imply that there exists an index $n_0$ such that for any two $n_1$ and $n_2$ both larger than $n_0$ , we have $n_1 < n_2 \Rightarrow a_{n_1} < a_{n_2}$ ?
Now I hope this statement is correct (and non-trivial!)

**emakarov** · May 31st 2012, 03:16 PM

What about $a_{2n}=1-1/n^2$ and $a_{2n+1}=1-1/n$ ?

**jens** · May 31st 2012, 04:58 PM

Ok, thanks!! So my conjecture was hopelessly wrong…

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