How to prove this sequence Cauchy

Question

Let $(a_{n})$ be a real Cauchy sequence with its elements and limit nonzero, then how can I show $(\frac{1}{a_{n}})$ is also Cauchy? Here is my try:

$|\frac{1}{a_{n}}-\frac{1}{a_{m}}|=|\frac{a_{m}-a_{n}}{a_{m}a_{n}}|=\frac{|a_{m}-a_{n}|}{|a_{m}a_{n}|}$

then for $|a_{m}-a_{n}|$, we can use the fact that $(a_{n})$ is Cauchy to restrict it, but what about $|a_{m}a_{n}|$? Although I tried to use the truth that $(a_{n})$ is bounded, but I find it can not give the right inequality to prove Cauchy, and I find that if we can prove $|a_{n}|$ has a lower bound, then we can finish the proof, but how can I prove that's true?

Answer 1

Hint: since $a_n \to L \ne 0$, there exists some $N\in \mathbb N$ such that:

$$\forall n\ge N: |a_n-L|<\frac {|L|}2$$

That is, $|a_n|>\dfrac {|L|}2$.

Answer 2

$\mathbb{R}$ is a complete metric space. i.e every Cauchy sequence is also a convergent sequence

So $a_{n}$ is a Cauchy sequence if and only if $a_{n}$ is a convergent sequence

Then if $\lim_{n \to \infty} a_{n} = L \not= 0 \implies \lim_{n \to \infty} \frac{1}{a_{n}} = \frac{1}{L} \implies \frac{1}{a_{n}}$

is a convergent sequence $\implies$ is a Cauchy sequence.

I hope help you