Properties of Convergent Sequences - Sum and Multiple Laws

# 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:

\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:

\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:

\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:

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