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- Thread starter Kamataat
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- #28

matt grime

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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?

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