Eventually self-outgrowing = eventually monotonic?

Hi,

Consider a bounded sequence $\displaystyle (a_n)_{n \in \mathbb{N}}$ whose terms are eventually "outgrown" by other terms of the sequence, i.e., for any index $\displaystyle n$ there exists an $\displaystyle n' > n$ such that $\displaystyle a_{n'} > a_n$. Does this property imply that $\displaystyle a_n$ is eventually monotonic, i.e., that there exists an index $\displaystyle n_0$ such that $\displaystyle a_n$ is monotonically increasing from index $\displaystyle n_0$ 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 $\displaystyle a_n = n + (-1)^n$

Thanks,

jens

Re: Eventually self-outgrowing = eventually monotonic?

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

Re: Eventually self-outgrowing = eventually monotonic?

Thanks for your counterexample!

Sorry for the mess, but I had only sequences in mind whose variations become "tiny", that is, whose $\displaystyle \lim\sup$ and $\displaystyle \lim\inf$ coincide. But I forgot to tell. So let me correct my question:

Assume $\displaystyle a_n$ converges to a finite limit and for any index $\displaystyle n$, there exists $\displaystyle n' > n$ such that $\displaystyle a_{n'} > a_n$. Does this imply that there exists an index $\displaystyle n_0$ such that for any two $\displaystyle n_1$ and $\displaystyle n_2$ both larger than $\displaystyle n_0$, we have $\displaystyle n_1 < n_2 \Rightarrow a_{n_1} < a_{n_2}$?

Now I hope this statement is correct (and non-trivial!)

Re: Eventually self-outgrowing = eventually monotonic?

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

Re: Eventually self-outgrowing = eventually monotonic?

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