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:
(1)
\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 $n ≥ N$. $\blacksquare$

We will now look at proving whether certain sequences are convergence or divergent.

Example 1

Prove that the sequence $\left ( \frac{2}{1} , \frac{3}{2} , \frac{4}{3} , ...\right )$ is convergent.

We note that this sequence can be rewritten to have its general term $a_n = \frac{n+1}{n}$ for $n \in \mathbb{N}$. We note that $a_1 = 2$, $a_2 = 1.5$, $a_3 = 1.33...$ and we suspect that this sequence is convergent namely to the real number $L = 1$ since from Calculus we know that $\lim_{n \to \infty} \frac{n+1}{n} = \lim_{n \to \infty} 1 + \frac{1}{n} = 1$. We will now formally prove this guess.

Let $\epsilon > 0$ be given. We need to a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - 1 \mid < \epsilon$. Now we note that:

(2)
\begin{align} \quad \mid a_n - 1 \mid = \biggr \rvert \frac{n+1}{n} - 1 \biggr \rvert = \biggr \rvert 1 - \frac{1}{n} -1 \biggr \rvert = \biggr \rvert \frac{1}{n} \biggr \rvert = \frac{1}{n} \end{align}

We now want to know whether or not $\frac{1}{n} < \epsilon$ for $n ≥ N$. We note that $\frac{1}{n} < \epsilon$ if and only if $n > \frac{1}{\epsilon}$. Now by the Archimedean property since $\frac{1}{\epsilon} > 0$ (since $\epsilon > 0$) then there exists a natural number $N$ such that $0 < \frac{1}{\epsilon} < N$ and thus $\frac{1}{\epsilon} < N$ which implies $\epsilon > \frac{1}{N}$, and $\frac{1}{N} > \frac{1}{n}$ since $n ≥ N$. Therefore $\mid a_n - 1 \mid = \frac{1}{n} < \epsilon$.

Example 2

Prove that the sequence $\left ( \frac{1}{n} \right )_{n=1}^{\infty}$ is convergent.

We note that $a_n = 1$, $a_2 = \frac{1}{2}$, $a_3 = \frac{1}{3}$, … so we suspect that this sequence converges to zero, that is $\lim_{n \to \infty} \frac{1}{n} = 0$.

Let $\epsilon > 0$ be given. We need to a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n \mid = \biggr \rvert \frac{1}{n} \biggr \rvert < \epsilon$. Now since $\epsilon > 0$ we have that $\frac{1}{\epsilon} > 0$, and once again by the Archimedean property we know that there exists a natural number $N$ (dependent on $\epsilon$ ) such that $\frac{1}{N} < \epsilon$, and so if $n ≥ N$, then $\frac{1}{n} ≤ \frac{1}{N} < \epsilon$. Therefore this sequence converges to zero.

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