Eventually self-outgrowing = eventually monotonic?

Hi,

Consider a bounded sequence whose terms are eventually "outgrown" by other terms of the sequence, i.e., for any index there exists an such that . Does this property imply that is eventually monotonic, i.e., that there exists an index such that is monotonically increasing from index 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

Thanks,

jens

Re: Eventually self-outgrowing = eventually monotonic?

and .

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 and coincide. But I forgot to tell. So let me correct my question:

Assume converges to a finite limit and for any index , there exists such that . Does this imply that there exists an index such that for any two and both larger than , we have ?

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

Re: Eventually self-outgrowing = eventually monotonic?

What about and ?

Re: Eventually self-outgrowing = eventually monotonic?

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