Properties of Convergent Sequences - Sum and Multiple Laws

We will now look at some very important properties of convergent sequences. 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 check out the Properties of Convergent Sequences - Product and Quotient 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 $(a_n + b_n)$ is convergent and $\lim_{n \to \infty} a_n + b_n = A + B$.

• Proof: Let $(a_n)$ and $(b_n)$ be convergent sequences such that $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$. We want to show that $\lim_{n \to \infty} (a_n + b_n) = A + B$, that is $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid (a_n + b_n) - (A + B) \mid < \epsilon$.

• We note by the triangle inequality that:

(1)

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

• Now since $\lim_{n \to \infty} a_n = A$, then for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists a natural number $N_1 \in \mathbb{N}$ such that if $n ≥ N_1$ then $\mid a_n - A \mid < \epsilon_1 = \frac{\epsilon}{2}$.

• Similarly since $\lim_{n \to \infty} b_n = B$, then for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists a natural number $N_2 \in \mathbb{N}$ such that if $n ≥ N_2$ then $\mid b_n - B \mid < \epsilon_2 = \frac{\epsilon}{2}$.

• Now we want both of these inequalities to hold, so choose $N = \mathrm{max} \{ N_1, N_2 \}$ and so therefore:

(2)

\begin{align} \quad \quad \mid (a_n + b_n) - (A + B) \mid = \mid (a_n - A) + (b_n - B) \mid ≤ \mid a_n - A \mid + \mid b_n - B \mid < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

• Therefore $\lim_{n \to \infty} (a_n + b_n) = A + B$. $\blacksquare$

Theorem 2: Let $(a_n)$ be a convergent sequence such that $\lim_{n \to \infty} a_n = A$, and let $k$ be any nonzero real number. Then the sequence $(ka_n)$ is convergent, and $\lim_{n \to \infty} ka_n = kA$.

• Proof: Let $(a_n)$ be a convergent sequence such that $\lim_{n \to \infty} a_n = A$, and let $k \in \mathbb{R}$ be nonzero. We want to show that $(ka_n)$ is convergent by showing that $\lim_{n \to \infty} ka_n = kA$, that is $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid ka_n - kA \mid < \epsilon$.

• We note that:

(3)

\begin{align} \quad \mid ka_n - kA \mid = \mid k (a_n - A) \mid = \mid k \mid \mid a_n - A \mid \end{align}

• Now since $(a_n)$ is a convergent sequence then for $\epsilon_1 = \frac{\epsilon}{\mid k \mid} > 0$ there exists an $N_1 \in \mathbb{N}$ such that if $n ≥ N_1$ then $\mid a_n - A \mid < \epsilon_1 = \frac{\epsilon}{\mid k \mid}$.

• We want the condition above to hold, so choose $N = N_1$. Therefore it follows that:

(4)

\begin{align} \quad \mid ka_n - kA \mid = \mid k (a_n - A) \mid = \mid k \mid \mid a_n - A \mid < \mid k \mid \epsilon_1 = \frac{\mid k \mid \epsilon}{\mid k \mid} = \epsilon \end{align}

• Therefore $\lim_{n \to \infty} ka_n = kA$. $\blacksquare$