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

**alice8675309** · May 9th 2011, 11:36 AM

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.

**Plato** · May 9th 2011, 11:43 AM

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?

**alice8675309** · May 9th 2011, 11:51 AM

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?

**Plato** · May 9th 2011, 11:55 AM

That does it. Right?

**alice8675309** · May 9th 2011, 11:57 AM

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?

**Plato** · May 9th 2011, 12:02 PM

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

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