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I am working over the review for my next exam coming up in Calculus, and am stuck on this problem:

Prove that, if $\lim_{n \to \infty}a_n=a$ and $a>0$, then there exists a positive integer $N$ such that $a_n>0$ for all $n>N$.

So, first I thought of using the property of a limit, that is, let $\epsilon > 0$, because the sequence $\{a_n\}$ converges to $a$, we are able to write $|a_n-a|<\epsilon$. However, I'm not sure where to go from there.

Now, the problem seems to want a proof using the Archimedean Property, however I am unsure of how exactly to begin applying it, maybe assume that $|a_n-a|\geq 0$, and thus, I could find an integer $N$ such that, for all $n>N$, $0 < \frac{1}{n} < |a_n-a|$. However, I don't see how to get there.

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2 Answers 2

up vote 1 down vote accepted

Since $\,a>0\,$ there exists $\,\epsilon>0\,\,\,s.t.\,\,\,a-\epsilon>0\,$ . Now, there exists $$N_\epsilon\in\Bbb N\,\,\,s.t.\,\,\,n>N_\epsilon\Longrightarrow |a_n-a|<\epsilon$$

But

$$|a_n-a|<\epsilon\Longleftrightarrow a-\epsilon<a_n<a+\epsilon$$

Well, wrap the above up and end the argument.

By the way, you can choose $\,\epsilon=a/2\,$, say.

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Thank you, that helps a lot. Because $a-\epsilon < a_n < a+\epsilon$ and $a-\epsilon>0$, we then know that, for all $n>N_\epsilon$ that $0<a_n$. Thus that concludes the proof, because we now know that for some $n>N$ that $a_n>0$! It took me a bit of re-reading your answer to fully get that ha! Thank you. –  Alex Oct 23 '12 at 4:13

Hint: Let $\epsilon=\dfrac{a}{2}$. By the definition of limit, there is an $n$ such that if $n\gt N$ then $|a_n-a|\lt \epsilon$. Conclude from this that if $n\gt N$ then $a_n\gt 0$.

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