Proof of Limit of a Sequence Raised to a Positive Integer Power

Problem statement:

If $\{a_n\}$ is a sequence of positive numbers such that $\lim\limits_{n \to \infty}a_n =L$, prove using $\epsilon, N$ definition\begin{align*}\lim\limits_{n \to \infty}a_n^p = L^p\end{align*} Where $p \in \mathbb{N}$

Solution:

We need to show that for any $\epsilon >0$ there exists some $N \in \mathbb{N}$ such that if $n>N$\begin{align*}|a_n^p-L^p| < \epsilon\end{align*}

observations:

Using the identity

\begin{align*}x^p-y^p = (x-y)(x^{p-1}+x^{p-2}y+...+xy^{p-2}+y^{p-1})\end{align*}

we have\begin{align*}|a_n^p -L^p| = |a_n-L||a_n^{p-1}+a_n^{p-2}L+...+a_nL^{p-2}+L^{p-1}|\\\implies |a_n^p -L^p| \leq |a_n-L|||a_n^{p-1}|+|a_n^{p-2}L|+...+|a_nL^{p-2}|+|L^{p-1}||\end{align*}

We need to bound $|a_n^kL^m|$ where $k,m \in \mathbb{N}$ and $0\leq k,m \leq p-1$ with $k+m =p-1$.\begin{align*}|a_n-L| <1 &\implies -1 < a_n-L<1 \implies L-1<a_n<L+1\\&\implies |a_n| < |L|+1 \qquad \text{ for some } n > N_1\\&\implies |a_n|^k |L|^m <(|L|+1)^k|L|^m\end{align*}Now, if $n > N_1$\begin{align}|a_n^p -L^p| < |a_n-L|\left((|L|+1)^{p-1}+(|L|+1)^{p-2}|L|+...+(|L|+1)|L|^{p-2}+|L|^{p-1} \right) \qquad (*)\end{align}Hence we bound $|a_n^p-L^p|$ above by $\epsilon$ by requiring that:\begin{align*}|a_n-L| < \frac{\epsilon}{\left((|L|+1)^{p-1}+(|L|+1)^{p-2}|L|+...+(|L|+1)|L|^{p-2}+|L|^{p-1} \right)}\end{align*}for some $p \in \mathbb{N}$.

proof:

Now, we can complete the proof. Because $a_n \to L$, $\exists N_1 \in \mathbb{N}$ such that if $n > N_0$ then for $\epsilon_0 = 1$,\begin{align*}|a_n-L| <1 &\implies |a_n| < |L|+1 \\&\implies |a_n|^k |L|^m <(|L|+1)^k|L|^m\end{align*}For $k,m \in \mathbb{N}$ and $0\leq k,m \leq p-1$ with $k+m =p-1$.

Again, using the fact that $a_n \to L$, $\exists N_1 \in \mathbb{N}$ such that if $n > N_1$ then for $p \in \mathbb{N}$ and some $\epsilon_1$,\begin{align*}|a_n-L| < \epsilon_1 = \frac{\epsilon}{\left((|L|+1)^{p-1}+(|L|+1)^{p-2}|L|+...+(|L|+1)|L|^{p-2}+|L|^{p-1} \right)}\end{align*}Then, for any arbitrary $\epsilon > 0$, there exists $N = max\{N_0, N_1\}$ such that if $n > N$, both the previous inequalities hold:\begin{align*}|a_n^p -L^p| = |a_n-L||a_n^{p-1}+a_n^{p-2}L+...+a_nL^{p-2}+L^{p-1}| \\\implies |a_n^p -L^p| \leq |a_n-L|\left||a_n^{p-1}|+|a_n^{p-2}L|+...+|a_nL^{p-2}|+|L^{p-1}|\right|\\\implies |a_n^p -L^p| < |a_n-L|\left((|L|+1)^{p-1}+(|L|+1)^{p-2}|L|+...+(|L|+1)|L|^{p-2}+|L|^{p-1} \right)\\ \implies |a_n^p -L^p| < \frac{\epsilon}{\left((|L|+1)^{p-1}+...+|L|^{p-1} \right)} * \left((|L|+1)^{p-1}+...+|L|^{p-1} \right) = \epsilon\end{align*}Hence, if $a_n\to L$, $a_n^p \to L^p$ for positive sequences of $a_n$ and $p \in \mathbb{N}$

Questions:

I have not found a proof of power of a sequence with limit on the internet. Hence I have a few concerns about the valadity of my proof:

• Is the general approach of the proof sound? are there any flaws in the general argument?
• Is the specific expression marked $(*)$ sound?
• Is the argument using $N_0$, $N_1$, $\epsilon_1$technically correct?

Would be interested in also getting more elegant $\epsilon, N$ proofs for this problem.

Edit: Removed redundant step