Thread: Limit Proof

I am having a bit of trouble with what should be a simple proof:

Let (an) be a convergent sequence of real numers, an >= 0 for all n in the natural number set. If the limit of an is a (an->a) show that a >= 0.

I understand that I need to use an epsilon neighbourhood to show a contradiction when a is < 0 but am having a hard time finding an epislon that produces a contradiction.

Thanks for the help

Suppose that a<0 then choose .
Because of sequence convergence we have

But that means: .

Do you see the contradiction?

I am sure I am just being very dense but where did the rhs of the inequality come from?

Originally Posted by jlindbo

I am sure I am just being very dense but where did the rhs of the inequality come from?

, so

since , we have

now continue to show the contradiction

Doh ;P

And since that produces the contradiction.

Thanks

Originally Posted by jlindbo

Doh ;P

And since that produces the contradiction.

Thanks

correct

This is a hugely important idea in sequence convergence that has a similar application to continuous functions.

If the limit of a sequence is positive then almost all the terms of that sequence must be positive. There is a similar statement of a negative limit.