Proving that a sequence $|a_n|\leq 1/n$ is Cauchy.

Question

I'm trying to prove that for any sequence $(a_n)$ where $|a_n| < 1/n$, $(a_n)$ is a Cauchy sequence.

To do it I'm using the fact that any sequence that converges is Cauchy. I am using the Squeeze theorem to show that $0 \leq |a_n|\leq 1/n$ and since $0 \rightarrow 0, 1/n \rightarrow 0$ it must be that $(|a_n|)$ converges to $0$ and therefore is Cauchy.

Is this a valid proof?

Answer 1

Yes that is valid, but perhaps not what the questioner had in mind. You can prove it is Cauchy without mentioning a limit at all. For instance, if $\epsilon > 0$ is given and $N \in \mathbb N$ satisfies $N > \frac{2}{\epsilon}$ then $$n,m \ge N \implies |a_n - a_m| \le |a_n| + |a_m| < \frac 1n + \frac 1m \le \frac 2N < \epsilon.$$

Answer 2

You know that every convergent sequence is a Cauchy sequence (it is immediate regarding to the definitions of both a Cauchy sequence and a convergent sequence).

Your demonstration of the convergence of the sequence is right, so yes, you have proven that $(\vert a_n\vert)$ is a Cauchy sequence.

Answer 3

Since $|a_n|$ converges to zero, $a_n$ must converge to zero as well.

Hence $a_n$ is Cauchy.