Limit Product/Quotient Laws for Convergent Sequences

We will now look at the limit product and quotient laws (law 3 and law 4 from the Limit of a Sequence page) and prove their validity.

Product Law for Convergent Sequences

Law 3 (Product Law of Convergent Sequences): If the limits of the sequences $\{ a_n \}$ and $\{ b_n \}$ are convergent, that is $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$, then $\lim_{n \to \infty} (a_nb_n) = \lim_{n \to \infty} a_n \cdot \lim_{n \to \infty} b_n = AB$.

• Proof of Law 3: Let $\{ a_n \}$ and $\{ b_n \}$ be convergent sequences. Since $\{ a_n \}$ is convergent, it is also bounded, that is there exists $M ≥ 0$ such that $\mid a_n \mid < M$. Now we note that $\lim_{n \to \infty} a_n = A$ implies that $\forall \epsilon > 0 \: \exists N_1 \in \mathbb{N}$ such that if $n ≥ N_1$ then $\mid a_n - A \mid < \frac{\epsilon}{\mathrm{max} \{ \mid b \mid, 1 \}} < \epsilon$. Similarly, $\lim_{n \to \infty} b_n = B$ implies that $\forall \epsilon > 0 \: \exists N_2 \in \mathbb{N}$ such that if $n ≥ N_2$ then $\mid b_n - B \mid < \frac{\epsilon}{2 \mathrm{max} \{M, 1\}} < \epsilon$. Note that these choices seem rather abstract, but will make more sense subsequently in the proof.

• Now we want to show that $\forall \epsilon > 0 \: \exists N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_nb_n - AB \mid < \epsilon$. Choose $N = \mathrm{max} \{ N_1, N_2 \}$ so that both $\mid a_n - A \mid < \frac{\epsilon}{\mathrm{max} \{ \mid b \mid, 1 \}}$ and $\mid b_n - B \mid < \frac{\epsilon}{2 \mathrm{max} \{M, 1\}}$ hold. With some algebraic manipulation and the triangle inequality, we obtain that:

(1)

\begin{align} \quad \mid a_nb_n - AB \mid = \mid a_n(b_n - B) + B(a_n - A) \mid ≤ \mid a_n \mid \mid b_n - B \mid + \mid B \mid \mid a_n - A \mid \end{align}

• And making the appropriate substitutions we get that:

(2)

\begin{align} \quad \mid a_nb_n - AB \mid ≤ \mid a_n \mid \mid b_n - B \mid + \mid B \mid \mid a_n - A \mid < M \frac{\epsilon}{2 \mathrm{max} \{M, 1\}} + \mid b \mid \frac{\epsilon}{2\mathrm{max} \{ \mid b \mid, 1 \}} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

• Therefore $\lim_{n \to \infty} (a_nb_n) = AB$. $\blacksquare$

Quotient Law for Convergent Sequences

Law 4 (Quotient Law of Convergent Sequences): If the limits of the sequences $\{ a_n \}$ and $\{ b_n \}$ are convergent, that is $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$, then $\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{\lim_{n \to \infty} a_n}{\lim_{n \to \infty} b_n} = \frac{A}{B}$ provided that $\lim_{n \to \infty} b_n ≠ 0$.

• Proof of Law 4: Let $\epsilon > 0$ be given. We want to shown that $\forall \epsilon > 0 \: \exists N \in \mathbb{N}$ such that if $n ≥ N$, then $\mid \frac{a_n}{b_n} - \frac{A}{B} \mid < \epsilon$. Doing some algebraic manipulation and using the triangle inequality, we get:

(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} + \frac{A}{b_n} - \frac{A}{B} \biggr \rvert = \biggr \rvert \frac{a_n - A}{b_n} + (B - b_n) \cdot \frac{A}{B \cdot b_n} \biggr \rvert ≤ \biggr \rvert \frac{a_n - A}{b_n} \biggr \rvert + \biggr \rvert (B - b_n) \cdot \frac{A}{B \cdot b_n} \biggr \rvert \end{align}

• Doing some more work, we obtain that:

(4)

\begin{align} \quad \quad \biggr \rvert \frac{a_n}{b_n} - \frac{A}{B} \biggr \rvert ≤ \frac{\mid a_n - A\mid }{\mid b_n \mid} + \biggr \rvert \frac{A}{B} \biggr \rvert \cdot \frac{\mid B - b_n \mid}{\mid b_n \mid} \end{align}

• We note that since $\{ b_n \}$ is convergent to the limit $B$, 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:

(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 + \biggr \rvert \frac{2A}{B^2} \biggr \rvert \cdot \mid B - b_n \mid \end{align}

• Now $\lim_{n \to \infty} a_n = A$ implies that $\forall \epsilon > 0 \: \exists N_1 \in \mathbb{N}$ such that if $n ≥ N_1$ then $\mid a_n + A \mid < \frac{\mid B \mid \epsilon}{4}$ to ensure that $\mid a_n - A \mid \cdot \frac{2}{\mid B \mid} < \frac{\epsilon}{2}$.

• Similarly, $\lim_{n \to \infty} b_n = B$ implies that $\forall \epsilon > 0 \: \exists N_2 \in \mathbb{N}$ such that if $n ≥ N_2$ then $\mid b_n + B \mid < \biggr \rvert \frac{B^2}{4A} \biggr \rvert \epsilon$ to ensure that $\mid b_n - B \mid \biggr \rvert \frac{2A}{B^2} \biggr \rvert < \frac{\epsilon}{2}$.

• Now choose $N = \mathrm{max} \{ N_1, N_2 \}$ so that $\mid a_n + A \mid < \frac{\mid B \mid \epsilon}{4} < \frac{\epsilon}{2}$ and $\mid b_n + B \mid < \biggr \rvert \frac{2A}{B^2} \biggr \rvert \epsilon < \frac{\epsilon}{2}$, and therefore:

(6)

\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 + \biggr \rvert \frac{2A}{B^2} \biggr \rvert \cdot \mid B - b_n \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

• Therefore for $n ≥ N$, $\biggr \rvert \frac{a_n}{b_n} - \frac{A}{B} \biggr \rvert < \epsilon$, so $\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{A}{B}$. $\blacksquare$