Proving $\lim_{n \to\infty} \frac{1}{n^p}=0$ for $p > 0$? I'm trying to prove 3.20a) from baby Rudin. We are dealing with sequences of real numbers.

Theorem.
$$\lim_{n \to {\infty}} \frac{1}{n^p} = 0; \hspace{30 pt}\mbox {$p > 0$}$$
Proof. Let $\epsilon > 0$. Because of the Archimedan property of real numbers, there exists an $N \in \Bbb{N}$ such that $n_0 \geq N$ implies $\frac{1}{n_0} < \epsilon$ and thus implies $n_0 > \frac{1}{\epsilon}$. Thus $\exists n \geq n_0 : n > (\frac{1}{\epsilon})^k$ where $k$ is any number. For the interesting case, pick $k = \frac{1}{p}$ where $p > 0$. Thus $n > (\frac{1}{\epsilon})^{\frac{1}{p}}$ implies $$\frac{1}{n} < \frac{1}{\left(\frac{1}{\epsilon}\right)^{\frac{1}{p}}} \Longrightarrow \frac{1}{n^p} < \epsilon$$
which further implies $d(n, N) < \epsilon$. QED.


Another approach: you can show $x^p \rightarrow \infty $ as $x\rightarrow \infty$. Rewrite as $$\frac {x^{p+1}}{x}$$ , which is an indeterminate $\infty/\infty$ , and use L'Hopital, to get $$\frac {(p+1)x^{p}}{1} $$. Since $p$ is fixed and $p+1>1$, you can show this goes to $\infty$ , and then $1/x^p\rightarrow 0$
