Additional Convergence Tests

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

• Section 11.6 – Absolute Convergence and the Ratio and Root Tests

• Section 11.7 – Strategy for Testing Series

Expected Educational Results

• Objective 23–1: I can identify and explain when a series is conditionally convergent.

• Objective 23–2: I can correctly cancel factorial factors.

• Objective 23–3: I can use the Ratio Test to determine if a series converges or diverges.

• Objective 23–4: I can use the Root 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.

Ratio Test

Definition: Ratio Test

• If ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|<1}$$\displaystyle\lim_{n\rightarrow\infty}\left|\,\frac{a_{n+1}}{a_n}\,\right|<1$, then the series $\sum |a_n|$$\sum{|\,a_n\,|}$ is absolutely convergent.

• If ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|>1}$$\displaystyle\lim_{n\rightarrow\infty}\left|\,\frac{a_{n+1}}{a_n}\,\right|>1$, then the series is divergent.

• If ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=1}$$\displaystyle\lim_{n\rightarrow\infty}\left|\,\frac{a_{n+1}}{a_n}\,\right|=1$, then the Ratio Test is inconclusive.

NOTE: The ratio test is helpful when the series contains factorials or exponential factors.

Investigation 02

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

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

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

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

4. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{(n-1)!}{2^n}}$$\displaystyle \sum_{n=0}^{\infty}{\frac{(n-1)!}{2^n}}$

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

6. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{{10}^n}{n{4}^{2n+1}}}$$\displaystyle \sum_{n=1}^{\infty}{\frac{10^n}{n4^{2n+1}}}$

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 02-1 *)
2
SumConvergence[(2n+1)!/(5^n*n!), 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 11.6: # 1, 7, 9, 11, 13, 15, 17, 19.

CC BY-NC-SA 4.0

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

Last Modified: Tuesday, 17 November 2020 6:42 EDT