I have the following exercise, for this I am not allowed to use natural logarithm of algebra of convergent sequences.
Exercise. Prove that $(n\in\mathbb{N}\mapsto 1/n^p)\longrightarrow 0$, i.e. the sequence $n\in\mathbb{N}\mapsto 1/n^p$ converges to $0$. The following is my proof.
Proof. For this proof we are going to use the Definition of convergence of a sequence, for brevity we define $s := (n\in\mathbb{N}\mapsto 1/n^p)$. Let $\varepsilon$ be a positive real number, we note that
\begin{align*} |s_n - 0|\lt\varepsilon &\iff \left|\frac{1}{n^p} - 0 \right|\lt\varepsilon &\text{because of $s$ was given}\\ &\iff \frac{1}{n^p} \lt\varepsilon\\ &\iff n^p>\frac{1}{\varepsilon}, \end{align*}
consequently,
\begin{align*} \tag{1} |s_n - 0|\lt\varepsilon &\iff n^p>\frac{1}{\varepsilon}. \end{align*}
Since $n,p,1/\varepsilon>0$ then $$n^p>\frac{1}{\varepsilon} \iff n>\sqrt[p]{\frac{1}{\varepsilon}},$$
wich, togheter with the afirmation at $(1)$, implies \begin{align*} \tag{2} |s_n - 0|\lt\varepsilon &\iff n>\sqrt[p]{\frac{1}{\varepsilon}}. \end{align*}
We define $N:= \lceil \sqrt[p]{\frac{1}{\varepsilon}} \rceil + 1$, we note that $N$ is a natural number and let $n$ be a natural number greater or equal to $N$. From how we declare $n$ it follows that $n>\sqrt[p]{\frac{1}{\varepsilon}}$, wich implies, because of the proposition at $(2)$, that $|s_n-0|\lt\varepsilon$. Therefore, $s\longrightarrow 0$.
I want to know if my proof is correct and what I can improve in it, thank you so much for your help.