Convergence and Divergence of Sequences

# Convergence and Divergence of Sequences

We will now look at two very important terms when it comes to categorizing sequences.

Definition: A sequence $(a_n)$ is said to be convergent to the real number $L$ and we write $\lim_{n \to \infty} a_n = L$ if $\forall \epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - L \mid < \epsilon$. If $\lim_{n \to \infty} a_n$ does not exist (that is the limit is infinity, negative infinity, or just doesn't converge in general) then we say that the sequence $(a_n)$ is divergent. |

From this definition of convergence, we immediately have the following theorem of equivalence statements.

Theorem 1: Let $(a_n)$ be a convergent sequence. Then the following statements are equivalent:1. The sequence $(a_n)$ converges to the real number $L$.2. For every $\epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - L \mid < \epsilon$.3. For every $\epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ the terms satisfy $L - \epsilon < a_n < L + \epsilon$.4. For every $\epsilon > 0$, for the $\epsilon$-neighbourhood $V_{\epsilon}(L)$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $x_n \in V_{\epsilon} (L)$. |

**Proof:**$1 \Leftrightarrow 2$ Suppose that the sequence $(a_n )$ converges to the real number $L$. Then $\lim_{n \to \infty} a_n = L$ and by the definition, for every $\epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - \mid < \epsilon$.

- $2 \Leftrightarrow 3$ Suppose that for every $\epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - L \mid < \epsilon$. Taking the last inequality, by an earlier theorem we note that $\mid a_n - L \mid < \epsilon$ if and only if:

\begin{align} -\epsilon < a_n - L < \epsilon \\ \Leftrightarrow L - \epsilon < a_n < L + \epsilon \end{align}

- $3 \Leftrightarrow 4$ If for every $\epsilon > 0$, for the $\epsilon$-neighbourhood $V_{\epsilon}(L)$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $x_n \in V_{\epsilon} (L)$, then $x_n \in V_{\epsilon} (L)$ for all $n ≥ N$. $\blacksquare$

The following theorem tells us that the sequence $\displaystyle{\{ \frac{1}{n} \right \}}_{n=1}^{\infty}$ converges to $0$.

Theorem 1: The sequence $\displaystyle{\{ \frac{1}{n} \right \}}_{n=1}^{\infty}$ converges to $0$. |

**Proof:**Let $\epsilon > 0$ be given. Observe that:

\begin{align} \quad \biggr \lvert \frac{1}{n} - 0 \biggr \rvert = \biggr \lvert \frac{1}{n} \biggr \rvert = \frac{1}{n} \end{align}

- We want $\frac{1}{n} < \epsilon$, i.e., $\frac{1}{\epsilon} < n$. So choose $N > \frac{1}{\epsilon}$. Then if $n \geq N$ we have that $n > \frac{1}{\epsilon}$, and so $\frac{1}{n} < \epsilon$, so from above:

\begin{align} \quad \biggr \vert \frac{1}{n} - 0 \biggr \rvert = \frac{1}{n} < \epsilon \end{align}

- Therefore $\displaystyle{\{ \frac{1}{n} \right \}}_{n=1}^{\infty}$ converges to $0$. $\blacksquare$