cauchy sequence prove

Question

I have a question about cauchy sequences. If we think $a_n$ is a cauchy sequence, we ca say $a_n^2$ is cauchy sequence too. But how can I prove it? And if we would say $a_n^2$ is cauchy sequence, can we say $a_n$ is a cauchy sequence? Thanks.

Answer 1

$(a_n)_n$ is Cauchy$\implies (a_n)_n $ bounded by $M>0$ for example. Let $\epsilon>0$ given. then $\exists N\geq 0 : \forall (p,q)\in \mathbb N^2$ $(p,q>N\implies|a_p-a_q|<\frac{\epsilon}{2M})$

and

$p,q>N\implies |a_p^2-a_q^2|=|a_p-a_q||a_p+a_q|\leq2M|a_p-a_q|<\epsilon$

Answer 2

If $(x_n)$ is a Cauchy sequence then

$$\forall \epsilon>0 \exists N\in\mathbb{N}: n,m\ge N\implies |x_n-x_m|<\epsilon.$$

We have that a Cauchy sequence is bounded. That is, there exists $M$ such that $|a_n|\le M,\forall n\in\mathbb{N}.$ Thus,

$$|x_n^2-x_m^2|=|x_n+x_m||x_n-x_m|\le 2M|x_n-x_m|,\forall n,m\in\mathbb{N}.$$ So

$$\forall \epsilon>0 \exists N\in\mathbb{N}: n,m\ge N\implies |x_n^2-x_m^2|<2M\epsilon.$$ That is, $(x_n^2)$ is a Cauchy sequence.

The converse is false. Just consider $x_n=(-1)^n.$ Note that $x_n^2=1$ (and thus a Cauchy sequence) while $x_n$ does not converge.

Answer 3

Since $\{a_n\}$ is a Cauchy sequence, it is limited. Let $M$ be a finite upper bound, than you have that for every $\varepsilon>0$ there exists $\bar{n}$ such that for every $n,m>\bar{n}$ $$ a_n^2-a_m^2=(a_n-a_m)*(a_n+a_m)<\varepsilon*2M. $$ Hence you can prove that $\{a_n^2\}$ is a Cauchy sequence too.

The second question is false. For example, let $$ a_n= \left\{ \begin{array}{ll} 1 & \mbox{ if n is even}\\ -1 & \mbox{ if n is odd} \end{array} \right. $$