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

• May 9th 2011, 11:36 AM
alice8675309
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 $\geqslant$k.

What I have:

Let $\varepsilon$>0. Then there is a k such that for all n $\geqslant$k, |x_n-a|< $\varepsilon$. Assume a>0. Then since $\exists$k such that ( $\forall$n $\geqslant$k, |x_n-a|< $\varepsilon$. We know that 0<|[x_n-a|< $\varepsilon$ which is a<|xn|< $\varepsilon$+a. Since a>0 we have |xn|>a>0.
• May 9th 2011, 11:43 AM
Plato
Quote:

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 $\geqslant$k.

Pick $\varepsilon = \frac{a}{2}$.
If $|x_n-a|<\frac{a}{2}$ what do get when rewritten without absolute values signs?
• May 9th 2011, 11:51 AM
alice8675309
Quote:

Originally Posted by Plato
Pick $\varepsilon = \frac{a}{2}$.
If $|x_n-a|<\frac{a}{2}$ what do get when rewritten without absolute values signs?

$\frac{a}{2}$<xn< $\frac{3a}{2}$ ? Which means that if $\frac{a}{2}$= $\varepsilon$ then xn> $\frac{a}{2}$= $\varepsilon$? so if xn> $\varepsilon$ and $\varepsilon$>0 so xn>0?
• May 9th 2011, 11:55 AM
Plato
That does it. Right?
• May 9th 2011, 11:57 AM
alice8675309
Quote:

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?
• May 9th 2011, 12:02 PM
Plato
There is really no way to answer that question, other than to say one word: experience.