# Convergent sequence sufficiency

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. –  Giovanni De Gaetano Nov 3 '12 at 14:29
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) –  Giovanni De Gaetano Nov 3 '12 at 15:02
I think you meant " the real sequence $\,\{1/2^n\}\,$ converges to zero. –  DonAntonio Nov 3 '12 at 16:50
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. –  Giovanni De Gaetano Nov 5 '12 at 10:18
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.) –  Did Nov 5 '12 at 11:24

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

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