Let . Suppose , then .
Let then choose
The is to get a new and then, for a we have
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.
No. A series doesnt have to go to 0 to be convergent. It only has to have a limit, meaning that when n -> infinite an -> x.
Convergent series - Wikipedia, the free encyclopedia here it can be seen as it is written in my book. It is stated around the 5th line or something like that.
Also, what is in my assignment doesnt state that it doesnt have a limit, it only states what i wrote in the beginning of my first post.
Im pretty sure that Abu-Khalil has the solution, i simply dont understand why he defines r the way he does in the middle.