Thread: Prove that there is a threshold k such that xn>0 for all n>=k.

Let (x_n) be a convergent sequence with limit a. Assume a>0. Prove that there is a threshold k such that x_n>0 for all n k.

What I have:

Let >0. Then there is a k such that for all n k, |x_n-a|< . Assume a>0. Then since k such that ( n k, |x_n-a|< . We know that 0<|[x_n-a|< which is a<|xn|< +a. Since a>0 we have |xn|>a>0.

Originally Posted by alice8675309

Let (x_n) be a convergent sequence with limit a. Assume a>0. Prove that there is a threshold k such that x_n>0 for all n k.

Pick .
If what do get when rewritten without absolute values signs?

Originally Posted by Plato

Pick .
If what do get when rewritten without absolute values signs?

<xn< ? Which means that if = then xn> = ? so if xn> and >0 so xn>0?

That does it. Right?

Originally Posted by Plato

That does it. Right?

yup. My question now is, when you look at a something to prove how do you know when to choose epsilon to be something other than just greater than 0 like you did in the problem by letting epsilon be a/2?

There is really no way to answer that question, other than to say one word: experience.