Sequences

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 11.1 – Sequences

Expected Educational Results

• Objective 18–01: I can classify a sequence as arithmetic, geometric, or fibonacci.

• Objective 18–02: I can explain what it mean for a sequence to converge and diverge.

• Objective 18–03: I can analytically (i.e., non-graphically and non-numerically) determine if a sequence converges or diverges.

• Objective 18–04: I can analytically (i.e., non-graphically and non-numerically) determine if a sequence is increasing or decreasing.

• Objective 18–05: I can analytically (i.e., non-graphically and non-numerically) determine if a sequence is bounded.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Sequences

Squeeze Theorem

Theorem: Squeeze Theorem

If $a_n\le b_n\le c_n$$a_n\leq b_n\leq c_n$ and ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}{c}_n=L}$$\displaystyle \lim_{n\to\infty}{a_n} = \lim_{n\to\infty}{c_n} = L$, then ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{b}_n=L}$$\displaystyle \lim_{n\to\infty}{b_n}=L$.

Theorem

If ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}|a_n|=0}$$\displaystyle \lim_{n\to\infty}\big|{a_n}\big| = 0$, then ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=0}$$\displaystyle \lim_{n\to\infty}{a_n}=0$.

Theorem

If ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=L}$$\displaystyle \lim_{n\to\infty}{a_n} = L$ and the function $f$$f$ is continuous at $L$$L$, then ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}f(a_n)=f(L)}$$\displaystyle \lim_{n\to\infty}{f\big(a_n\big)}=f(L)$.

Investigation 12

1. Use ${\displaystyle b_n=\frac{1}{n}}$$\displaystyle b_n=\frac{1}{n}$, $n\ge 1$$n\geq 1$, and ${\displaystyle c_n=-\frac{1}{n}}$$\displaystyle c_n=-\frac{1}{n}$, $n\ge 1$$n\geq 1$, to determine if ${\displaystyle a_n=\frac{\sin (n)}{n}}$$\displaystyle a_n=\frac{\sin{(n)}}{n}$, $n\ge 1$$n\geq 1$, converges or diverges. Explain.

2. Use ${\displaystyle b_n=\frac{1}{\sqrt{n}}}$$\displaystyle b_n=\frac{1}{\sqrt{n}}$, $n\ge 1$$n\geq 1$, and ${\displaystyle c_n=-\frac{1}{\sqrt{n}}}$$\displaystyle c_n=-\frac{1}{\sqrt{n}}$, $n\ge 1$$n\geq 1$, to determine if ${\displaystyle a_n=\frac{\cos (n)}{\sqrt{n}}}$$\displaystyle a_n=\frac{\cos{(n)}}{\sqrt{n}}$, $n\ge 1$$n\geq 1$, converges or diverges. Explain.

Investigation 13

Determine if the following sequences converge or diverge. Explain.

1. ${\displaystyle a_n=(-1{)}^n\frac{1}{n^2+2}}$$\displaystyle a_n=(-1)^n\frac{1}{n^2+2}$, $n\ge 0$$n\geq 0$.

2. ${\displaystyle a_n=(-1{)}^{n+1}\frac{n}{n+2}}$$\displaystyle a_n=(-1)^{n+1}\frac{n}{n+2}$, $n\ge 0$$n\geq 0$.

3. ${\displaystyle a_n=(-1{)}^n\frac{n}{n^2+2}}$$\displaystyle a_n=(-1)^n\frac{n}{n^2+2}$, $n\ge 0$$n\geq 0$.

4. ${\displaystyle a_n=\cos (x)\sqrt{x}}$$\displaystyle a_n=\cos{(x)}\sqrt{x}$, $n\ge 0$$n\geq 0$.

Homework

At this time, you should be able to complete the following assignments:

• Section 11.1: # 73, 75, 77.

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Created: Wednesday, 27 October 2021 - 07:51 (EDT)