# Cauchy convergent sequences

Suppose that $(a_n)$ and $(b_n)$ are convergent sequences and that $b_n > 0$ for all $n$. Is it true that $(a_n / b_n)$ is Cauchy? If it is true, prove it. If it is not true, give a counterexample to show why it is not true.

My approach: let $lim(a_n)$ = $L$ and $\lim(b_n)$ = $M$, with $M \neq 0$. Since $(a_n)$ and $(b_n)$ converge, then by the property of limits $(a_n / b_n)$ will converge too with $M \neq 0$.

I don't know how to make it so that it is Cauchy.

Put $a_n = 1/\sqrt{n}$ and $b_n = 1/n$. You have $a_n/b_n = \sqrt{n}$.

• Ah, you beat me; I was thinking of $a_n=1$, $b_n=1/n$. – user99680 Nov 6 '13 at 1:59

Not necessarily. Take for example, $a_n=\frac{(-1)^n}{n}$, and $b_n=\frac{1}{n}$. Then $a_n/b_n=(-1)^n$, which is not Cauchy.

The limit law you've cited is correct when $M \ne 0$, so try a case when $M = 0$. In particular, take $a_n$ to be a constant sequence (e.g. $a_n = 1$ for all $n$) and $b_n = \frac{1}{n}$, perhaps. Then

$$\frac{a_n}{b_n} = \frac{1}{\frac{1}{n}} = n$$

is an unbounded, and thus divergent sequence.

• @OrchidFibio Yes, that's correct. That's why the limit law doesn't work if $M \ne 0$. – user61527 Nov 6 '13 at 2:03
• @OrchidFibio Take $a_n = 1/n$ and $b_n = 1/n$. Then $a_n / b_n$ is a Cauchy sequence. – user61527 Nov 6 '13 at 2:06
• @OrchidFibio The limit law states that if $M \ne 0$, then the limit is $L/M$. It doesn't say anything at all about the case $M = 0$, and can't help you conclude that a sequence is divergent. It's also not true that $(a_n)$ and $(b_n)$ are subsequences of $(a_n / b_n)$. In this case, just simply take $a_n$ to be a constant sequence (i.e. $a_n = 1$) and $b_n = 1/n$; then $a_n / b_n = n$ is a divergent sequence. – user61527 Nov 6 '13 at 2:16