Properties of Cauchy Sequences - Product and Quotient Laws

Recall from the Cauchy Sequences of Real Numbers page that a sequence $(a_n)$ of real numbers is said to be Cauchy if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:

(1)

\begin{align} \quad |a_m - a_n| < \epsilon \end{align}

We will now prove that the product of two Cauchy sequences is also a Cauchy sequence.

Theorem 1: Let $(a_n)$ and $(b_n)$ be Cauchy sequences. Then $(a_nb_n)$ is a Cauchy sequence.

• Proof: Let $(a_n)$ and $(b_n)$ be Cauchy sequences. We want to show that $(a_nb_n)$ is also a Cauchy sequence, that is $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n ≥ N$ then $\mid a_nb_n - a_mb_m \mid < \epsilon$. Doing some algebraic manipulation we get that:

(2)

\begin{align} \quad \quad \mid a_nb_n - a_mb_m \mid = \mid a_nb_n - a_mb_n + a_mb_n - a_mb_m \mid = \mid b_n(a_n - a_m) + a_m(b_n - b_m) \mid < \mid b_n \mid \mid a_n - a_m \mid + \mid a_m \mid \mid b_n - b_m \mid \end{align}

• Now recall that every Cauchy sequence is bounded. Since $(a_n)$ and $(b_n)$ are Cauchy sequences it follows that for $0 < M_1, M_2 \in \mathbb{R}$ that $\mid a_n \mid < M_1$ and $\mid b_n \mid < M_2$ for all $n \in \mathbb{N}$ and so:

(3)

\begin{align} \quad \quad \quad \mid a_nb_n - a_mb_m \mid = ... < \mid b_n \mid \mid a_n - a_m \mid + \mid a_m \mid \mid b_n - b_m \mid < M_2 \mid a_n - a_m \mid + M_1 \mid b_n - b_m \mid \end{align}

• Also, since $(a_n)$ is Cauchy there exists an $N_1 \in \mathbb{N}$ such that if $m, n ≥ N_1$ then $\mid a_n - a_m \mid < \frac{\epsilon}{2M_2}$.

• Similarly since $(b_n)$ is Cauchy there exists an $N_2 \in \mathbb{N}$ such that if $m, n ≥ N_2$ then $\mid b_n - b_m \mid < \frac{\epsilon}{2M_1}$.

• Choose $N = \mathrm{max} \{ N_1, N_2 \}$ and so for $m, n ≥ N$:

(4)

\begin{align} \quad \quad \mid a_nb_n - a_mb_m \mid = ... < M_2 \mid a_n - a_m \mid + M_1 \mid b_n - b_m \mid < M_2 \frac{\epsilon}{2M_2} + M_1 \frac{\epsilon}{2M_1} = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

• Therefore $(a_nb_n)$ is a Cauchy sequence. $\blacksquare$

Observe that if $(a_n)$ and $(b_n)$ are Cauchy sequences then it need not be that $\left ( \frac{a_n}{b_n} \right )$ is a Cauchy sequence. For example, consider the following sequences:

(5)

\begin{align} \quad (a_n) = \left ( \frac{1}{n} \right ) \quad , \quad (b_n) = \left ( \frac{1}{n^2} \right ) \end{align}

These sequences both converge (to $0$) and are hence Cauchy by the Cauchy convergence criterion. However the following sequence is not Cauchy:

(6)

\begin{align} \quad \left ( \frac{a_n}{b_n} \right ) = \left ( \frac{\frac{1}{n}}{\frac{1}{n^2}} \right) = (n) \end{align}

This is because $(n)$ is divergent.

However, if $(b_n)$ is Cauchy and does not converge to $0$ then everything works out.

Theorem 2: Let $(a_n)$ and $(b_n)$ be Cauchy sequences and suppose that $(b_n)$ does not converge to $0$. Then $\left ( \frac{a_n}{b_n} \right )$ is a Cauchy sequence.

• Proof: Since $(a_n)$ and $(b_n)$ are Cauchy they are both convergent sequences by the Cauchy convergence criterion and since $(b_n)$ does not converge to $0$ we have that $\displaystyle{\left ( \frac{a_n}{b_n} \right)}$ converges. But again by the Cauchy convergence criterion we have that then $\displaystyle{\left ( \frac{a_n}{b_n} \right)}$ is a Cauchy sequence. $\blacksquare$