Comparison Tests

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

• Section 11.4 – The Comparison Tests

Expected Educational Results

• Objective 20–01: I can analytically (i.e., not graphically and not numerically) verify all conditions needed to use a Comparison Test for a given series.

• Objective 20–02: I can use the Limit Comparison Test to determine if a series converges or diverges.

• Objective 20–03: I can use the Comparison Test to determine if a series converges or diverges.

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.

Comparison Tests

Limit Comparison Test

Definition: Limit Comparison Test

Suppose $\sum a_n$$\sum{a_n}$ and $\sum b_n$$\sum{b_n}$ are series with positive terms. If ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=c}$$\displaystyle\lim_{n\to\infty}{\frac{a_n}{b_n}}=c$ where $c$$c$ is a finite number and $c>0$$c>0$, then either both series converge or both series diverge.

Explanation of Limit Comparison Test

The Limit Comparison Test is written as a conditional statement.

1. “Suppose $\sum a_n$$\sum{a_n}$ and $\sum b_n$$\sum{b_n}$ are series with positive terms” is a requirement in order to use the Limit Comparison Test.

   1. This requirement states all $a_n$$a_n$ must be positive and all $b_n$$b_n$ must be positive.

2. “If ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=c}$$\displaystyle\lim_{n\to\infty}{\frac{a_n}{b_n}}=c$ where $c$$c$ is a finite number and $c>0$$c>0$” is called the hypothesis of the conditional statement. You need to verify the hypothesis is true:

   1. The limit must be a finite number, i.e., the limit must exist, i.e., the limit must be a Real number;

   2. The must be a positive Real number.

3. “then either both series converge or both series diverge” is called the conclusion of the conditional statement.

   1. The conclusion is True ONLY IF the hypothesis is True. That is either

      1. both $\sum a_n$$\sum{a_n}$ and $\sum b_n$$\sum{b_n}$ converge; or

      2. both $\sum a_n$$\sum{a_n}$ and $\sum b_n$$\sum{b_n}$ diverge.

NOTE: You must analytically (i.e., not graphically and not numerically) verify $a_n>0$$a_n>0$ and $b_n>0$$b_n>0$ before using the comparison tests. You must show work or an explanation on all Assessments.

NOTE: The Limit Comparison Test is typically much easier than the Comparison Test.

Example 01:

Determine if ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{n^2+3n+5}{n^4+n^2+2}}}$$\displaystyle \sum_{n=1}^{\infty}{\dfrac{n^2+3n+5}{n^4+n^2+2}}$ is convergent or divergent.

Solution:

Let $a_n={\displaystyle \frac{3n^2+3n+5}{2n^4+n^2+2}}$$a_n=\dfrac{3n^2+3n+5}{2n^4+n^2+2}$.

We need to find a valid $b_n$$b_n$:

• As $n$$n$ gets large, the numerator is dominated by the leading term, $3n^2$$3n^2$.

• As $n$$n$ gets large, the denominator is dominated by the leading term, $2n^4$$2n^4$.

• Use the above information to obtain a possible $n^{\text{th}}$$n^{\text{th}}$-term for $b_n$$b_n$: ${\displaystyle b_n={\displaystyle \frac{3n^2}{2n^4}}={\displaystyle \frac{3}{2n^2}}}$$\displaystyle b_n=\dfrac{3n^2}{2n^4}=\dfrac{3}{2n^2}$.

• We must be able to determine if ${\displaystyle \sum b_n=\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{3}{2n^2}}}$$\displaystyle \sum{b_n}=\sum_{n=1}^{\infty}{\dfrac{3}{2n^2}}$ converges or diverges:

  • ${\displaystyle \sum b_n}$$\displaystyle \sum{b_n}$ is a $p$$p$-series, with $p=2$$p=2$ where $p>1$$p>1$, so, ${\displaystyle \sum b_n}$$\displaystyle \sum{b_n}$ converges.

Check the requirements for the Limit Comparison Test:

• Since $n>1$$n>1$ and $a_n$$a_n$ consists of addition, multiplication, and division of positive values, $a_n>0$$a_n>0$ for all $n$$n$;

• Since $n>1$$n>1$ and $b_n$$b_n$ consists of multiplication and division of positive values, $b_n>0$$b_n>0$ for all $n$$n$;

Verify the hypothesis is True:

• ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\frac{3n^2+3n+5}{2n^4+n^2+2}}{\frac{3}{2n^2}}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{3n^2+3n+5}{2n^4+n^2+2}}\left({\displaystyle \frac{2n^2}{3}}\right)=1\in {\mathbb{R}}^{+}}$$\displaystyle\lim_{n\to\infty}{\frac{a_n}{b_n}}=\lim_{n\to\infty}{\dfrac{\frac{3n^2+3n+5}{2n^4+n^2+2}}{\frac{3}{2n^2}}}=\lim_{n\to\infty}{\dfrac{3n^2+3n+5}{2n^4+n^2+2}\left(\dfrac{2n^2}{3}\right)}=1\in\mathbb{R}^+$

Make a conclusion:

• Since the hypothesis is True and ${\displaystyle \sum b_n}$$\displaystyle \sum{b_n}$ converges, then ${\displaystyle \sum a_n}$$\displaystyle \sum{a_n}$ converges.

Investigation 01

Determine if the following series are convergent or divergent. Explain.

1. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{1}{2\sqrt{n}+1}}$$\displaystyle \sum_{n=0}^{\infty}{\frac{1}{2\sqrt{n}+1}}$

2. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{1}{n^2+5n+7}}$$\displaystyle \sum_{n=0}^{\infty}{\frac{1}{n^2+5n+7}}$

3. ${\displaystyle \sum _{n=2}^{\mathrm{\infty }}\frac{n^3+3n^2+2n+4}{n^4-1}}$$\displaystyle \sum_{n=2}^{\infty}{\frac{n^3+3n^2+2n+4}{n^4-1}}$

4. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\sqrt{n^3+2}}}$$\displaystyle \sum_{n=1}^{\infty}{\frac{1}{n\sqrt{n^3+2}}}$

5. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{(n+1{)}^2(n-1{)}^3}{n(n+4{)}^5}}$$\displaystyle \sum_{n=1}^{\infty}{\frac{(n+1)^2(n-1)^3}{n(n+4)^5}}$

6. ${\displaystyle \sum _{n=3}^{\mathrm{\infty }}\frac{(n+3{)}^2}{(n-2{)}^4}}$$\displaystyle \sum_{n=3}^{\infty}{\frac{(n+3)^2}{(n-2)^4}}$

7. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{2^n}{3^n+5}}$$\displaystyle \sum_{n=0}^{\infty}{\frac{2^n}{3^n+5}}$

8. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{3^n}{2^n+5n}}$$\displaystyle \sum_{n=0}^{\infty}{\frac{3^n}{2^n+5n}}$

Use Technology to Verify Convergence or Divergence of a Series

NOTE: On all Assessments, you may use technology to verify your series convergence/divergence. You must show all Calculus II related work.

Mathematica

1
(* Example from Investigation 01-1 *)
2
SumConvergence[1/(2Sqrt[n]+1), n]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: http://www.jjw3.com/TECH_Common_Functions.pdf.

2. SumConvergence[ ] has at least two arguments:

   1. The $n^{\text{th}}$$n^{\text{th}}$-term of the series;

   2. $n$$n$ where $n$$n$ is the index variable of the series.

3. For help on using the SumConvergence[ ] function:

   1. In Mathematica, execute the code: $\text{?SumConvergence}$$\text{?SumConvergence}$

   2. Click on $\vee$$\or$ near the bottom-left of output

   3. Click on local

   4. Read how to use the SumConvergence[ ] function – you will be able to copy-paste code.

4. Remember, correct Mathematica code will be all black except for variables.

5. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

6. Mathematica will return “True” (series converges) or “False” (series diverges).

Homework

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

• Section 11.4: # 1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29.

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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License [http://creativecommons.org/licenses/by-nc-sa/4.0/].

Created: Tuesday, 17 November 2020 6:03 EDT Last Modified: Monday, 20 June 2022 - 11:10 (EDT)