- #1
shoescreen
- 15
- 0
Hello all,
I have been thinking of a way to prove divergence of a sequence that should work, but can't move past one road block.
Here's the idea, given a sequence a_n, say that i know for any consecutive numbers in the sequence, |a_(n+1) - a_(n)| > d, where d is a constant.
Now this alone should be enough to prove divergence, since the numbers cannot get any closer than distance d. Yet showing this is proving to be difficult for me (I'm sure someone reading this will get it in no time).
Oh and I am trying to prove this rigorously, that is for any constant A, there exists e > 0 such that for every n* there exists m > n* such that |a_n - A| > (or equal to) e.
The trick should be take e < d/2, and showing that for any value n* that satisfies |a_n* - A| < e, then |a_(n*+1) - A| cannot be less than e. It is this last statement that I am having a hard time reaching.
Please help me with this last step! I think a proof like this would be a pretty slick way of proving a sequence diverges.
Thanks
I have been thinking of a way to prove divergence of a sequence that should work, but can't move past one road block.
Here's the idea, given a sequence a_n, say that i know for any consecutive numbers in the sequence, |a_(n+1) - a_(n)| > d, where d is a constant.
Now this alone should be enough to prove divergence, since the numbers cannot get any closer than distance d. Yet showing this is proving to be difficult for me (I'm sure someone reading this will get it in no time).
Oh and I am trying to prove this rigorously, that is for any constant A, there exists e > 0 such that for every n* there exists m > n* such that |a_n - A| > (or equal to) e.
The trick should be take e < d/2, and showing that for any value n* that satisfies |a_n* - A| < e, then |a_(n*+1) - A| cannot be less than e. It is this last statement that I am having a hard time reaching.
Please help me with this last step! I think a proof like this would be a pretty slick way of proving a sequence diverges.
Thanks