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Definition: a real sequence is a mapping $f : N^+ \mapsto R$

The real sequence $\frac{1}{2^n}$ converges to $0$. If $\epsilon$ in $R^+$ is given, then $\left| \frac{1}{2^n} \right| < \epsilon$ when $2^n > \frac{1}{\epsilon}$. Because $2^n > n$ when $n \geq 1$, it is sufficient that $n > \frac{1}{\epsilon}$.

The above came from my textbook. I do not understand why $n > \frac{1}{\epsilon}$ is sufficient because $2^n > n$ when $n \geq 1$. What is the reason for that?

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    I guess your textbook is simply saying that $2^n > n > \frac{1}{\epsilon}$. So if $n> \frac{1}{\epsilon}$ also $2^n$ is.2012-11-03
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    I don't think that's what they're saying, since they're starting from 2^n > 1/e and going to n > 1/e2012-11-03
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    Your last implication is clearly false: $ 2^n > \frac{1}{\epsilon} $ does NOT imply $n> \frac{1}{\epsilon}$. Just consider $n=2$ and $\epsilon = \frac{1}{3}$. But it really does appear in your question that your book claims the right direction, if you were careful in reporting it. Can you check exactly what your textbook says please? (and perhaps, if you are not sure, quote it straightfully here)2012-11-03
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    I think you meant " the real **sequence** $\,\{1/2^n\}\,$ converges to zero.2012-11-03
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    @DonAntonio Sorry. I had to translate it to English from Dutch, I gave the definition to avoid any doubts.2012-11-05
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    @GiovanniDeGaetano I'm not implying it, the book is. That is why I am confused. Also, this is exactly what my book said although there may be some translation mistakes.2012-11-05
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    The sentence: "If $\epsilon \in \mathbb{R}^+$ is given, then $|\frac{1}{2^n}|< \epsilon$ when $2^n > \frac{1}{\epsilon}$" is perfectly fine to me. Is it also for you? And also the next one: "Because $2^n>n$ when $n\geq 1$, it is sufficient that $n> \frac{1}{\epsilon}$ [To have $2^n> \frac{1}{\epsilon}]$]" (where into the bracket [...] I put the implied part of the sentence) looks fine to me. Because of the reason explained in my first comment. I'm sorry but I really don't see the problem.2012-11-05
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    @GiovanniDeGaetano The second statement doesn't look fine to me. I know why you think it's fine, and if that was the case I would agree with you, but I don't think that's the case. Why would they even want to prove that $2^n > 1/e$ with the last sentence? They've proven that before because the inequality $\left| \frac{1}{2^n} \right| < \epsilon$ implies it. However, they haven't proven why $n > 1/e$, so my guess is that they're trying to get to that.2012-11-05
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    @GiovanniDeGaetano If you disagree, then how would you prove that $n > \frac{1}{\epsilon}$ ?2012-11-05
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    xcrypt: *they're starting from 2^n > 1/e and going to n > 1/e*... Not in the sense you believe they are. In fact, they show that, **IF** n > 1/e, **THEN** 2^n > 1/e. (Note the "it is sufficient". Example: To die, it is sufficient to jump from an airplane with no parachute. Hence "jump from plane with no parachute" $\implies$ "death", but people die of other causes as well, right?) So, if indeed they are "going from" something to something, it is in the other direction than the one you said. And their direction is a true implication. (To rehash what @GiovanniDeGaetano already said.)2012-11-05
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    @did All right then. I was never introduced to "it is sufficient" in mathematics before.2012-11-05

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Perhaps if we see it written in mathematics it will be clearer. Let us list down what we know:

$$(1)\;\;\;\;\;\;\;\forall\,n\in\Bbb N\,\,,\,2^n>n$$

$$(2)\;\;\;\;\;\;\;\;\forall\,\,\epsilon>0\,\,\,,\,\,\,\frac{1}{2^n}<\epsilon\Longleftrightarrow 2^n>\frac{1}{\epsilon}$$

$$(3)\;\;\;\;\;\;\;\text{Thus, if}\,\,n>\frac{1}{\epsilon}\,\,,\,\text{ we get:}$$

$$2^n\stackrel{\text{by}\,\,(1)}>n\stackrel{\text{if (3)}}>\frac{1}{\epsilon}\stackrel{\text{by (2)}}\Longrightarrow 2^n>\frac{1}{\epsilon}\Longleftrightarrow \frac{1}{2^n}<\epsilon$$

and then we're done.

Note that in the right arrow $\,\Longrightarrow\,$ in the last line above, we used transitivity of the relation $\,x>y\,$:

$$x>y>z\Longrightarrow x>z$$

Finally, how do we prove that there exists some $\,n\in\Bbb N\,$ that fulfills (3) above, no matter what $\,\epsilon\,$ , and thus what $\,1/\epsilon\,$ , is?

Very simple: this is the Archimedean Property of the natural numbers, and you can read about it here