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

1. ## 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$\displaystyle \geqslant$k.

What I have:

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

2. 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$\displaystyle \geqslant$k.
Pick $\displaystyle \varepsilon = \frac{a}{2}$.
If $\displaystyle |x_n-a|<\frac{a}{2}$ what do get when rewritten without absolute values signs?

3. Originally Posted by Plato
Pick $\displaystyle \varepsilon = \frac{a}{2}$.
If $\displaystyle |x_n-a|<\frac{a}{2}$ what do get when rewritten without absolute values signs?
$\displaystyle \frac{a}{2}$<xn<$\displaystyle \frac{3a}{2}$ ? Which means that if $\displaystyle \frac{a}{2}$=$\displaystyle \varepsilon$ then xn>$\displaystyle \frac{a}{2}$=$\displaystyle \varepsilon$? so if xn>$\displaystyle \varepsilon$ and $\displaystyle \varepsilon$>0 so xn>0?

4. That does it. Right?

5. 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?

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