Quotient of two Cauchy sequences

Question

So I recently was asked whether the quotient of two Cauchy sequences have to be Cauchy too. I said not, because it is easy to make a counterexample with like $a_{n} = \frac{1}{n}, b_{n} = \frac{1}{n^{2}}$. But my friend didn't think of any sequences less than $1$, and convinced himself it was true as a result, which was wrong under that question's context.

But his misunderstanding made me wonder, if you have two Cauchy sequences in some subset of the real numbers, $\{a_{n}\} , \{b_{n}\}$ (where the quotient is in the subset too) and you set that every term of both sequences is at least $1$, then is it provable that the quotient is Cauchy too?

I think it should be true, since I don't know how you would build a counterexample at all, but I don't know how to prove it. I tried supposing that the quotient was not Cauchy and then getting a contradiction with the Cauchy definitions of either sequence, but didn't get anywhere.

Answer 1

This is true: if $a_n$ and $b_n$ are Cauchy, then they are convergent (by completeness of $\mathbb{R}$). Now the condition that $a_n, b_n \ge 1$ insures that their limits are greater than or equal to $1$ (say $a_n \rightarrow a \ge 1$ and $b_n \rightarrow b \ge 1$). Now by the quotient rule $\frac{a_n}{b_n} \rightarrow \frac{a}{b} \in \mathbb{R}$ so $\frac{a_n}{b_n}$ is convergent which by completeness implies it is Cauchy. Hope this helps!

The reason your counterexample works is because of the undefined nature of $\frac{0}{0}$. Once you take that possiblility away things work a lot nicer!

Answer 2

The reals are complete, so Cauchy sequences are precisely convergent sequences.

Therefore we wish to prove the following:

The limit of a quotient is the quotient of the limits. That is, if $(a_n), (b_n)$ are sequences converging to $a, b$ respectively, where $b \not = 0$, then $\frac{a_n}{b_n} \to \frac{a}{b}$.

This follows from the following two lemmas:

The limit of a product is the product of the limits. That is, if $a_n \to a, b_n \to b$, then $a_n b_n \to a b$.

Proof: Let $n$ be big enough that $|b_n - b| \leq \frac{\epsilon}{|a|+1}$, $|a_n - a| \leq \epsilon / |b|$, and $|a_n| \leq |a|+1$. Then $$|a_n b_n - a b| = |a_n (b_n - b) + b (a_n - a)| \leq \underbrace{|a_n|}_{\leq |a|+1} \underbrace{|(b_n - b)|}_{\leq \frac{\epsilon}{|a|+1}} + |b| \underbrace{|(a_n - a)|}_{\leq \frac{\epsilon}{|b|}}$$ which is bounded by a constant multiple of $\epsilon$.

The limit of a reciprocal is the reciprocal of the limit. That is, if $a_n \to a$ and $a \not = 0$, then $\frac{1}{a_n} \to \frac{1}{a}$.

Proof: Let $\epsilon$ be fixed, and pick $n$ such that $|a_n - a| \leq \epsilon$. Then $$\left | \frac{1}{a_n} - \frac{1}{a} \right| = \left| \frac{a-a_n}{a_n a} \right| \leq \frac{1}{|a|} \epsilon \frac{1}{|a_n|}$$ We may assume $\epsilon$ is smaller than $\frac{|a|}{2}$, so $|a_n - a| \leq \frac{|a|}{2}$, so $|a_n| \leq \frac{|a|}{2}$, so $$\left| \frac{1}{a_n} - \frac{1}{a} \right| \leq \frac{2 \epsilon}{|a|^2}$$