Properties of Convergent Sequences - Product and Quotient Laws

# Properties of Convergent Sequences - Product and Quotient Laws

We will now look at some more very important properties of convergent sequences. Once again, it is stressed that these properties always hold for convergent sequences, so be sure not to assume they hold for divergent sequences or sequences whose convergence/divergence is unknown.

Also be sure to see the Properties of Convergent Sequences - Sum and Multiple Laws

 Theorem 1: Let $(a_n)$ and $(b_n)$ be convergent sequences such that $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$. Then the sequence $(a_nb_n)$ is convergent, and $\lim_{n \to \infty} a_nb_n = AB$.
• Proof: Let $(a_n)$ and $(b_n)$ be convergent sequences such that $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$. We want to show that $\lim_{n \to \infty} a_nb_n = AB$, that is, $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_nb_n - AB \mid < \epsilon$.
• We note with a little algebraic manipulation, the use of the triangle inequality, and noting that since $(b_n)$ is convergent, it must also be bounded and so for some positive real number $R$, $\mid b_n \mid < R$, we have that:
(1)
\begin{align} \quad \quad \mid a_nb_n - AB \mid = \mid (a_n - A)b_n + A(b_n - B) \mid ≤ \mid b_n \mid \mid a_n - A \mid + \mid A \mid \mid b_n - B \mid < R \mid a_n - A \mid + \mid A \mid \mid b_n - B \mid \end{align}
• Now since $(a_n)$ is convergent, then for $\epsilon_1 = \frac{\epsilon}{2R} > 0$ there exists a natural number $N_1 \in \mathbb{N}$ such that if $n ≥ N_1$ then $\mid a_n - A \mid < \epsilon_1 = \frac{\epsilon}{2R}$.
• Similarly since $(b_n)$ is convergent, then for $\epsilon_2 = \frac{\epsilon}{2(\mid A \mid + 1)} > 0$ there exists a natural number $N_2 \in \mathbb{N}$ such that if $n ≥ N_2$ then $\mid b_n - B \mid < \epsilon_2 = \frac{\epsilon}{2 (\mid A \mid + 1)}$.
• We want both of these conditions to hold, so choose $N = \mathrm{max} \{ N_1, N_2 \}$. Therefore:
(2)
\begin{align} \quad \quad \mid a_nb_n - AB \mid = ... < R \mid a_n - A \mid + \mid A \mid \mid b_n - B \mid < R \cdot \epsilon_1 + \mid A \mid \epsilon_2 = R \frac{\epsilon}{2R} + \mid A \mid \frac{\epsilon}{2 (\mid A \mid + 1)} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
• Therefore $\lim_{n \to \infty} a_nb_n = AB$. $\blacksquare$
 Theorem 2: Let $(a_n)$ and $(b_n)$ be convergent sequences such that $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$ and $B \neq 0$. Then the sequence $\left(\frac{a_n}{b_n} \right)$ is convergent, and $\lim_{n \to \infty} \left(\frac{a_n}{b_n} \right) = \frac{A}{B}$.
• Proof: Let $(a_n)$ and $(b_n)$ be convergent sequences such that $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$ where $B \neq 0$. We want to show that $\lim_{n \to \infty} \left(\frac{a_n}{b_n} \right) = \frac{A}{B}$, that is $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\biggr \rvert \frac{a_n}{b_n} - \frac{A}{B} \biggr \rvert < \epsilon$.
• With some algebraic manipulation:
(3)
\begin{align} \quad \quad \biggr \rvert \frac{a_n}{b_n} - \frac{A}{B} \biggr \rvert = \biggr \rvert \frac{a_n}{b_n} - \frac{A}{b_n} + \frac{A}{b_n} - \frac{A}{B} \biggr \rvert = \biggr \rvert \frac{a_n - A}{b_n} + (B - b_n) \cdot \frac{A}{B b_n} \biggr \rvert ≤ \biggr \rvert \frac{1}{b_n} \biggr \rvert \mid a_n - A \mid + \biggr \rvert \frac{A}{B b_n} \biggr \rvert \mid B - b_n \mid \end{align}
• Now since $(b_n)$ converges, then at some point, $\mid b_n \mid ≥ \frac{\mid B \mid}{2}$ which implies that $\frac{1}{\mid b_n \mid} ≤ \frac{2}{\mid B \mid}$. Therefore:
(4)
\begin{align} \quad \quad \biggr \rvert \frac{a_n}{b_n} - \frac{A}{B} \biggr \rvert = . . . ≤ \biggr \rvert \frac{1}{b_n} \biggr \rvert \mid a_n - A \mid + \biggr \rvert \frac{A}{B b_n} \biggr \rvert \mid B - b_n \mid < \frac{2}{\mid B \mid} \mid a_n - A \mid + \frac{2 \mid A \mid}{\mid B \mid^2} \mid B - b_n \mid \end{align}
• Now since $(a_n)$ converges, then for some $\epsilon_1 = \frac{\mid B \mid \epsilon}{4} > 0$ there exists an $N_1 \in \mathbb{N}$ such that if $n ≥ N_1$ then $\mid a_n - A \mid < \epsilon_1 = \frac{\mid B \mid \epsilon}{4}$.
• Similarly since $(b_n)$ converges, then for some $\epsilon_2 = \frac{\mid B \mid^2}{4(\mid A \mid +1)} \epsilon > 0$ there exists an $N_2 \in \mathbb{N}$ such that if $n ≥ N_2$ then $\mid b_n - B \mid < \epsilon_2 = \frac{\mid B \mid^2}{4(\mid A \mid +1)} \epsilon$.
• We want both of the conditions above to hold, so choose $N = \mathrm{max} \{ N_1, N_2 \}$ and so:
(5)
\begin{align} \quad \quad \biggr \rvert \frac{a_n}{b_n} - \frac{A}{B} \biggr \rvert = . . . < \frac{2}{\mid B \mid} \mid a_n - A \mid + \frac{2 \mid A \mid}{\mid B \mid^2} \mid B - b_n \mid < \frac{2}{\mid B \mid} \epsilon_1 + \frac{2 \mid A \mid}{\mid B \mid^2} \epsilon_2 = \frac{2}{\mid B \mid} \frac{\mid B \mid \epsilon}{4} + \frac{2 \mid A \mid}{\mid B \mid^2} \frac{\mid B \mid^2}{4(\mid A \mid +1)} \epsilon < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
• Therefore $\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{A}{B}$. $\blacksquare$