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Section 9.1 Sequences

Definition 9.1.1.

A sequence is an infinite list of numbers:

\begin{equation*} a, \quad a_1, \quad a_2, \quad a_3, \dots \underbrace{a_n}_{\text{n}^\text{th} \text{term, or general term}} \end{equation*}
\begin{equation*} 1, \quad 4, \quad 9, \quad 16, \quad 25, \dots, n^2 \end{equation*}

We can think of the previous example ( \(a_n = n^2\) ) as a function:

\begin{equation*} a(n) = n^2 \end{equation*}

These functions are writen in one of two ways:

  • recursive -- uses the previous term to define the following term

  • closed form -- non-recursive

Brace Notation.

  • \(\displaystyle \frac{1}{2}, \quad \frac{2}{5}, \quad \frac{7}{10}, \quad \frac{15}{17} \dots\)

  • \(\displaystyle a_n = \dfrac{2^n - 1}{n^2 + 1}\)

\begin{equation*} \left\{ \dfrac{2^n - 1}{n^2 + 1} \right\}_{n = 1}^{\infty} \end{equation*}

The sequence \(a_1, a_2, \dots, a_n\) has a limit, \(L\text{,}\) written:

\begin{equation*} \lim_{n\to\infty} a_n = L \end{equation*}

if \(a_n\) can be made as close to \(L\) as we like by taking n large enough.

  • If such \(L\) exists, we say the sequence converges to \(L\)

  • If it doesn't, we say the sequence diverged.

  • Suppose:

    \begin{align*} \amp \left\{ a_n \right\}_{n = 1}^{\infty} \text{ converges to } L_1 \\ \amp \left\{ b_n \right\}_{n = 1}^{\infty} \text{ converges to } L_2\\ \amp c \text{ is a constant} \end{align*}

    • \(\displaystyle \displaystyle \lim_{n\to\infty} c = c\)

    • \(\displaystyle \displaystyle \lim_{n\to\infty} ca_n = c \left( \displaystyle \lim_{n\to\infty} a_n \right) = cL_1\)

    • \(\displaystyle \displaystyle \lim_{n\to\infty} \left( a_n \pm b_n \right) = L_1 + L_2\)

    • \(\displaystyle \displaystyle \lim_{n\to\infty} \left( a_n b_n \right) = L_1 L_2\)

    • \(\displaystyle \displaystyle \lim_{n\to\infty} \dfrac{a_n}{b_n} = \dfrac{L_1}{L_2} \text{ provided } L_2 \neq 0\)

General Sequence Facts.

  • \(\displaystyle a_n = x^n \begin{cases} \text{converges to } 0 \text{ if } \mid x \mid \lt 1\\ \text{diverges if } \mid x \mid \gt 1 \end{cases}\)

  • \(\displaystyle a_n = \frac{1}{n^P} \begin{cases} \text{converges to } 0 \text{ if } P \gt 0\\ \text{diverges if } P \lt 0 \end{cases}\)

Example 9.1.1.

Consider \(a_n = \frac{n}{e^n}\) and determine if it converges or diverges:

\(\displaystyle \lim_{n\to\infty} \frac{n}{e^n} = \frac{\infty}{\infty} \text{ indeterminate form}\)

When we use \(x\) the variable is considered continuous (not skipping) but \(n\) is descrete (only integrers). Since \(n\) is not continuous, we can't use L'Hopital's Rule directly. We have to extend the domain from \(n\) to \(x\text{.}\)

\(\displaystyle \lim_{x\to\infty} \frac{x}{e^x} = \displaystyle \lim_{x\to\infty} \frac{1}{e^x} = \displaystyle \lim_{x\to\infty} \frac{1}{\infty} = 0\)

This will always equal to \(\displaystyle \lim_{n\to\infty} \frac{n}{e^n}\text{.}\)

We simply write:

\begin{equation*} \lim_{n\to\infty} \frac{n}{e^n} = \lim_{n\to\infty} \frac{1}{e^n} = 0 \end{equation*}

but know we skipped steps.