Difference between revisions of "Sequence"

Revision as of 16:50, 27 June 2022

A sequence is an ordered list of terms. Sequences may be either finite or infinite.

Definition

A sequence of real numbers is simply a function $f : \mathbb{N} \rightarrow \mathbb{R}$ . For instance, the function $f(x) = x^2$ defined on $\mathbb{N}$ corresponds to the sequence $X = (x_n) = (0, 1, 4, 9, 16, \ldots)$ .

Convergence

Intuitively, a sequence converges if its terms approach a particular number.

Formally, a sequence $(x_n)$ of reals converges to $L \in \mathbb{R}$ if and only if for all positive reals $\epsilon$ , there exists a positive integer $k$ such that for all integers $n \ge k$ , we have $|x_n - L| < \epsilon$ . If $(x_n)$ converges to $L$ , $L$ is called the limit of $(x_n)$ and is written $\lim_{n \to \infty} x_n$ . The statement that $(x_n)$ converges to $L$ can be written as $(x_n)\rightarrow L$ .

A classic example of convergence would be to show that $1/n\to 0$ as $n\to \infty$ .

Claim: $\lim_{n\to\infty}\frac{1}{n}=0$ .

Proof: Let $\epsilon>0$ be arbitrary and choose $N>\frac{1}{\epsilon}$ . Then for $n\ge N$ we see that

$n>\frac{1}{\epsilon}\implies \frac{1}{n}<\epsilon\implies \left|\frac{1}{n}-0\right|<\epsilon$

which proves that $|x_n-L|<\epsilon$ , so $1/n\to 0$ as $n\to \infty$ $\square$

Monotone Sequences

Many significant sequences have their terms continually increasing, such as $(n^2)$ , or continually decreasing, such as $(1/n)$ . This motivates the following definitions:

A sequence $(p_n)$ of reals is said to be

increasing if $p_n\leq p_{n+1}$ for all $n\in\mathbb{N}$ and strictly increasing if $p_n<p_{n+1}$ for all $n\in\mathbb{N}$ ,
decreasing if $p_n\geq p_{n+1}$ for all $n\in\mathbb{N}$ and strictly decreasing if $p_n>p_{n+1}$ for all $n\in\mathbb{N}$ ,
monotone if it is either decreasing or increasing.

Resources

Online Encyclopedia of Integer Sequences

@@ Line 11: / Line 11: @@
 A classic example of convergence would be to show that <math>1/n\to 0</math> as <math>n\to \infty</math>.
-'''Claim''': Let <math>\lim_{n\to\infty}\frac{1}{n}=0</math>.
+'''Claim''': <math>\lim_{n\to\infty}\frac{1}{n}=0</math>.
 ''Proof'': Let <math>\epsilon>0</math> be arbitrary and choose <math>N>\frac{1}{\epsilon}</math>.  Then for <math>n\ge N</math> we see that

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