Proof of $p \in \mathbb{R}_{+}$ then $\lim_{n \to \infty} \frac{1}{n^{p}} = 0$

Question

I want to show the following:

Let $p \in \mathbb{R}_{+}$, then $\lim_{n \to \infty} \frac{1}{n^{p}} = 0 $.

My proof is the following.

From the Archimedean property of real numbers $\forall p > 0$, $\exists m \in \mathbb{N}$ such that $m>p$. Then: \begin{equation} 0 < \frac{1}{n^{p}} = \left(\frac{1}{n}\right)^{p} < \left(\frac{1}{n}\right)^{m} \end{equation}

Now assume that: $\left(\frac{1}{n}\right)^{m} > \epsilon $ $\forall n \in \mathbb{N}$. Then it follows that:

$n <\frac{1}{\epsilon^{m}}$ for all $n \in \mathbb{N}$.

But this means that $\mathbb{N}$ is bounded above and this is absurd. Then $\exists n_{0}\in \mathbb{N}$ such that $\frac{1}{n^{p}} < \epsilon$, $\forall n > n_{0}$.

I'm not sure about the logical steps involved in the proof. Can you tell me if there is something wrong with it?

Thanks!

Answer 1

For p$\ge$1, n<nᵖ for all natural numbers n $\ge$ 2. So, 0<1/nᵖ<1/n , now by Archimeadian property, for any $\epsilon$ > 0, 0<1/n< $\epsilon$ for all n$\ge$N ,for some natural number N. Hence, 1/nᵖ <$\epsilon$ for all n$\ge$ N.
For 0<p<1, if possible let, 1/nᵖ > G for some positive real G, holds for all natural numbers n, then n<1/$G^{1/p}$ , but this is a contradiction to archimeadian property because Archimeadian property says that for any real x>0, there always exist a natural number n such that n>x. So, our assumption is wrong . Hence , 1/nᵖ <$\epsilon$ for all n$\ge$ N(for some natural number N).
So,for every $p \in \mathbb{R}_{+}$ , $\lim_{n \to \infty} \frac{1}{n^{p}} = 0$

Answer 2

Let $a_n=\frac{1}{n^p}$, we need to show that $\operatorname{lim}a_n=0$. For given $\epsilon> 0$. Let us choose $N\in \Bbb{N}$ such that $N >( \frac{1}{\epsilon})^\frac{1}{p}$(by the Archimedean property)so $N>(\frac{1}{\epsilon})^\frac{1}{p}$ for $p\geq 1$. Then, if$n \geq N$. Then, $n\geq N$ implies $\frac{1}{n}\leq\frac{1}{N}<\epsilon^\frac{1}{p}$. Consequently, if $n\geq N$, then

$$|a_n-0|=|\frac{1}{n^p}-0|=\frac{1}{n^p}<\epsilon$$ for $p\geq 1$.