I'll give you the sequence and it's limit from the new definition, you show that it 'converges' to that limit:
Hint: Take for every
After doing that, see if you can do it for any sequence.
I've been thinking in circles over this problem so I thought I'd give this forum a try - I hope someone can help.
I've been given a definition:
For all epsilon>0 , there exists a natural number N such that if |Xn-L| < epsilon, then n is greater than or equal to N.
I need to either find an infinite sequence Xn which converges but does not satisfy the conditon above OR
find a sequence Xn and L which do satisfy the condition above but Xn does not converge to L.
I'd be grateful if anybody could talk me through what features such a sequence should have.
Thanks for your reply, although I still can't get my head around it.
Does this sequence guarantee when but does not converge?
Would this show that does not guarantee ?
If the latter, does that apply for all possible epsilon greater than zero?
Ah, okay we have N=1 so n will always be greater than, or equal to it.
I guess my problem is, how does this work for all possible values of epsilon? e.g what if epsilon = 1/2, i.e. for n=1,
and how does this converge to 0 when n goes to infinity?
I appreciate your patience but I'm still missing something here.