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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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    @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