Question

Suppose that {a_{n}} is a Cauchy sequence. Prove that {(a_{n})^2} is a Cauchy sequence.

Answer 1

\(\{a_n\}\) is a Cauchy sequence, which means there exists an \(N\) such that \(m,n\ge N\) implies \(|a_m-a_n|<\epsilon\) for a given \(\epsilon>0\).

Since it is a Cauchy sequence, you also know that it's bounded (if you don't, it can be proved).

Using the fact that it's bounded, you know there exists a \(K\) for which \(|a_n|\le K\) for all \(n\).

To show that \(\{{a_n}^2\}\) is also a Cauchy sequence, you have to show that there is some \(N\) such that \(m,n\ge N\) implies
\[|{a_m}^2-{a_n}^2|<\epsilon\]
Factor this as a difference of squares and use the triangle inequality, and you get
\[|{a_m}^2-{a_n}^2|=|a_m-a_n||a_m+a_n|\le|a_m-a_n|\big(|a_m|+|a_n|\big)\]
Now you can use the fact that the sequence \(\{a_n\}\) is bounded to deal with the \(|a_m|+|a_n|\) factor.