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$\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.

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?

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?

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.