Prove that $\lim_{n\to\infty} 1/n^\alpha = 0$ for any $\alpha > 0$.

Question

My lecturer gave us this as exercise. Prove that $\lim_{n\to\infty} 1/n^\alpha = 0$ for any $\alpha > 0$.(using Archimedean principle and definition of convergence)

For $\varepsilon >0$, there exists $M \in N$ such that $|\frac 1{n^\alpha} - 0 |< \varepsilon$. I don't know how to proceed from here, and when I should use the Archimedean property.

Thank you in advance!!

Answer 1

$|\frac 1{n^\alpha} - 0 |< \varepsilon \iff n> \frac{1}{\varepsilon^{1/ \alpha}}$.

Answer 2

Attempt:

Let $\epsilon >0$ be given, and let

$M>0$, real, be such that $M>1/\epsilon.$

Archimedes :

There is a $n_0 \in \mathbb{Z^+}$ such that

$n_0 >M^{1/\alpha}.$

For $n \ge n_0:$

$|1/n^{\alpha}| \le |1/n_0^{\alpha}| \lt 1/M \lt \epsilon.$

Used: $f(x) = x^{\alpha}$ is an increasing function.

Answer 3

Can I try? Here's my attempt: Given $\varepsilon > 0$ by the Archimedean principle, $\exists N \in \mathbb{N}$, such that $0< \frac{1}{N} < \varepsilon$, and $\forall n \geq N \,\,, 0<\frac{1}{n} <\frac{1}{N} < \varepsilon $. Exponentiating with $\alpha > 0$, $0<(\frac{1}{n})^\alpha <(\frac{1}{N})^\alpha < (\varepsilon)^\alpha$. Since $\varepsilon$ was arbitrary, then we've shown that $\forall \varepsilon>0$, there exists $N \in \mathbb{N}$ such that $\forall n \geq N$, we have $|(\tfrac{1}{n})^\alpha -0|<\varepsilon$.