# Sequence Convergence and Divergence Proofs

Recall that a sequence $\{ a_n \}$ is said to converge to $L$ if $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - L \mid < \epsilon$. We also noted that a sequence diverges if it does not converge.

We will now look at some examples regarding the convergence / divergence of sequences.

## Example 1

**Prove or disprove the following statement: If the sequence $\{ \mid a_n \mid \}$ converges then the sequence $\{ a_n \}$ converges.**

Consider the logically equivalent contrapositive which says that if $\{ a_n \}$ diverges then $\{ \mid a_n \mid \}$ diverges. This statement is not true though. Consider the sequence $\{ (-1)^n \} = \{ -1, 1, -1, 1, ... \}$ which diverges. However, $\{ \mid (-1)^n \mid \} = \{ 1 \} = \{ 1, 1, ... \}$ converges.

Thus the statement is false in general.

## Example 2

**Prove or disprove the following statement: If $\{ a_n \}$ and $\{ b_n \}$ both diverge, then $\{ a_n b_n \}$ must diverge.**

Consider the sequences $\{ a_n \} = \{ (-1)^n \}$ and $\{ b_n \} = \{ (-1)^n \}$. Both of these sequences diverge, however, their product $\{ a_n b_n \} = \{ (-1)^n(-1)^n \} = \{ (-1)^{2n} \} = \{ 1 \}$ converges.

Therefore the statement is false in general.

## Example 3

**Prove or disprove the following statement: If $\{ a_n \}$ diverges to infinity and $\{ b_n \}$ converges to $L > 0$ then $\{ a_nb_n \}$ diverges to infinity.**

Suppose that $\{ b_n \}$ converges to $L > 0$. Then $\forall \epsilon > 0$ we have that if $n ≥ N_1$ then $\mid a_n - L \mid < \epsilon$. Now since $\{ b_n \}$ converges to $L > 0$ we also have that $\{ b_n \}$ is bounded. Let $B > 0$ be an upper bound to $\{ b_n \}$, that is, $b_n < B$ for all $n \in \mathbb{N}$.

Also suppose that $\{ a_n \}$ diverges to infinity, that is, $\lim_{n \to \infty} a_n = \infty$. Then for some natural number $N_2$ we have that if $n ≥ N_2$ then $a_n > \frac{M}{B}$.

We want to show that $\{ a_nb_n \}$ diverges to infinity, that is, $\lim_{n \to \infty} a_nb_n$. In other words, we want to show that for all $M \in \mathbb{R}$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $a_nb_n > M$.

Let $N = \max \{ N_1, N_2 \}$. Then if $n ≥ N$ we have that $a_nb_n > \frac{M}{B} b_n > \frac{M}{B} \cdot B = M$, so $\lim_{n \to \infty} a_nb_N = \infty$.

## Example 4

**Prove or disprove the following statement: If $\{ a_n \}$ diverges to infinity and $\{ b_n \}$ diverges to negative infinity then $\{ a_n + b_n \}$ converges to zero.**

Consider the sequence $\{ a_n \} = \{ n^{100} \}$ and the sequence $\{ b_n \} = \{ -n \}$. Notice that $\lim_{n \to \infty} a_n = \infty$ and $\lim_{n \to \infty} -n = -\infty$.

Now $\{ a_n + b_n \} = \{ n^{100} - n \}$ and $\lim_{n \to \infty} (n^{100} - n) = \infty \neq 0$.

Therefore the statement is false in general.