Limit of a sequence

  • Thread starter Kamataat
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  • #26
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Despite that I was able to follow that last example using many random values for e, I still do not understand why we have to make n_0 dependent on e. Since e a completly random positive number that gives us n_0, then why not do away with e and just pick n_0 > 0 randomly?

- Kamataat
 
  • #27
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Ok, e is COMPLETELY arbitary. But can the dependence of n_0 on e, n_0(e), also be arbitarily chosen (as long as we make sure that n_0 is positive)?

- Kamataat
 
  • #28
matt grime
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I really don't get those questions in this sense.

the e is very necessary in this definition of convergence, you cannot do away with e.

Also you do not "choose e" as your last post implicitly states.

n_0 may not be picked to be any number, there is a condition it must satisfy, that all terms after the n_0'th are no more than e away in distance. So it clearly will depend on e. If you did understand the example, where I pick n_0 to be ceiling(1/e) then you'd have surely not asked that question.

I mean suppose e were 1/10,000 then I can't pick n_0 arbitrarily and independently of e can I? say I set n_0 =1, then it is not true that for all n>1 that |1/n - 0| < 1/10,000

eg pick n any number from 2 to 10,0000.


The form of convergence proofs is: given any e>0, we must find an n_0 such that...

quantify it: for all e>0 there exists n_0 in N such that...

what would you suggest doing instead?
 
  • #29
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After thikig about it for a day, I guess I'll try some examples from my textbook. It seems clearer now. Anyway, thanks for all your time! I won't know whether I understand limits until I've tried some excercise problems, but I know your help made it a lot clearer to me!

- Kamataat
 

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