Hey people.

I have a series {an}.

I know that for every positive R > 0 exists an n so |an| >= R, where n is a natural number.

Now i have to show that {an} is divergent.

I dont exacly know how to do this. I tried to make a proof by conflict, saying that {an} was convergent and then choosing an R > epsilon + |a| where a is the limit of {an}. Then i thought i might be able to end up with something of the following:

For every R > 0 there is an n so |an| >= R. If R = epsilon + |a| -> |an| > epsilon + |a| -> |an| - |a| > epsilon..

Then i didnt know what to do, because i dont know if i can say that |an|-|a| =< |an - a| and then say that |an - a| >= |an| - |a| > epsilon -> |an - a| > epsilon.

Then it conflicts with the definition of convergence, cause it states that for any epsilon, there exists an N so |an - a| <= epsilon, which in this case isnt true.

Can i do this?

Thanks for the help.