if {${a_n}$} is a Cauchy sequence with ${a_n} \neq 0$ then {${1/a_n}$} is also a Cauchy sequence.

Question

New to this but I am not the best with proofs, in particularly, Cauchy sequences. This might be an easy proof and I am just over complicating it.

Question: Show that if {${a_n}$} is a Cauchy sequence with ${a_n} \neq 0$ for all $n \in N$, then {${1/a_n}$} is also a Cauchy sequence.

Answer 1

For the Cauchy sequence {$a_n$} you will also need to assume that $\lim_n a_n=l\ne 0$. Then $\lim_n \frac{1}{a_n}=\frac{1}{l}$.

Given $\epsilon>0$, there is a positive integer $N$ such that, for all $n,m>N$, $|a_n-a_m|<\epsilon$ and $|a_n|>\frac{|l|}{2}$. But then $$|\frac {1}{a_n}-\frac {1}{a_m}|=|\frac {a_n-a_m}{a_na_m}|<\frac {4\epsilon}{l^2}$$

Therefore {$\frac{1}{a_n}$} is a Cauchy sequence.

Answer 2

In general, for $A\subseteq \mathbb R$, and any Cauchy sequence $(a_n)\subseteq A$ and continuous function $f:\overline A \to \mathbb R$, we have that $(f(a_n))$ is a Cauchy sequence. That's because $(a_n)$ converges; $a_n\to a \in \overline A$, so $f(a_n)\to f(a)$ since $f$ is continuous. As a result, $f(a_n)$ is a Cauchy sequence, since it converges.

Now here is the problem: the above is only valid when $f(a)$ is defined. In your case, $f(x)=1/x$, $A=\{x\in\mathbb R|x>0\}$ so if $a_n\to 0$, $f(a)$ is undefined. So the result is only true when $a_n$ does not converge to $0$.