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Use only Archimedean Property of $\mathbb{R}$ to give a direct $\epsilon$-$N$ verification for $\lim\limits_{n\rightarrow\infty}\frac{1}{\sqrt{n}}=0$

Let $a_n=\{\frac{1}{\sqrt{n}}\}$ and $\epsilon>0$. Consider that there is a $\epsilon'=\epsilon^2$, then by the Archimedean Property, for every $\epsilon'$ that there exists an in $N$ such that $1/N<\epsilon'$ for all $n\geq N$. Then we have: $$\frac{1}{n}\leq\frac{1}{N}<\epsilon'\Rightarrow\frac{1}{\sqrt{n}}\leq\frac{1}{\sqrt{N}}<\sqrt{\epsilon'}=\epsilon$$

And then we have $\vert1/\sqrt{n}-0\vert=1/\sqrt{n}<\epsilon$; this shows that $a_n$ is converged to $0$ and gives $\lim\limits_{n\rightarrow\infty}\frac{1}{\sqrt{n}}=0$.

Since the question is asking a direct verification, but seems I am writing a proof. Could anyone give me a suggestion to change my answer to a direct verification ? Thanks

Simple
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1 Answers1

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I think that the verification in this case is analogous to a proof. However im going to point out some details of your proof which is ok just to make it more precise.

You first take your $\epsilon > 0$ which is now fixed (but arbitrary). Take as you said $\epsilon' = \epsilon^2$ For that given $\epsilon'$ by the archimedian property, there exist $N\in \Bbb{N}$ such that $1/N < \epsilon'$. (full stop)

Now for every $n \geq N$ we have that

$1/n \leq 1/N < \epsilon'$ which implies that $1/\sqrt{n} \leq 1/\sqrt{N} < \sqrt{\epsilon'}=\epsilon$

Therefore, for that given $\epsilon$ (which you chose arbitrarily), you found an $N \in \Bbb{N}$ such that for every $n \geq N$, $|1/\sqrt{n}-0|<\epsilon$ which is by definition that $1/\sqrt{n} \to 0$

Joaquin Liniado
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