$ a_n $ is a Cauchy sequence $ \Rightarrow $ $ a_n^2 $ is a Cauch sequence proof verification

Question

I want to prove that if $ a_n $ is a Cauchy sequence, then $ a_n^2 $ is also a Cauchy sequence, is my proof valid?

Let $ a_n $ be a Cauchy sequence. Therefore, $ a_n $ is bounded. Thus, there exists $ M \gt 0 $ such that for all $ n \in \mathbb{N}, \frac{-M}{2} \lt a_n \lt \frac{M}{2} $. Therefore, for all $ n, m \in \mathbb{N}, -M \lt a_n + a_m \lt M $.

Given $ \epsilon \gt 0 $, there exists $ N \in \mathbb{Z}^+ $, such that for all $ n, m \gt N, \, \left| a_n - a_m \right| \lt \frac{\epsilon}{M} $. Therefore, for all $ n, m \gt N $:

$ \left| a_n^2 - a_m^2 \right| = \left| a_n - a_m \right|\left|a_n + a_m\right| \lt \epsilon \cdot M = \epsilon$

This question is not about proving for the product of two Cauchy sequences, it specifically asks about proving the Square of a Cauchy sequence is also a Cauchy sequence

Answer 1

Just to follow all your steps, I guess in the last inequality it should be like $\frac{\varepsilon}{M} \cdot M = M$, otherwise it is correct.