Series

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

• Section 11.2 – Series

Expected Educational Results

• Objective 19–01: I can explain the difference between a series and a sequence.

• Objective 19–02: I can classify a series as geometric, $p$$p$-series, harmonic, or telescoping.

• Objective 19–03: I can explain what it means for a series to converge and diverge.

• Objective 19–04: I can explain why ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=0}$$\displaystyle\lim_{n\rightarrow\infty}{a_n}=0$ is a necessary but not sufficient condition for series convergence.

• Objective 19–05: I can use the appropriate Series Convergence Test to determine if a series converges or diverges.

• Objective 19–06: I can compute the interval on which a power series converges.

• Objective 19–07: I can compute the sum of a convergent geometric series.

• Objective 19–08: I can compute the sum of a telescoping series.

• Objective 19–09: I can use a geometric series to rewrite a repeating decimal as a rational number.

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.

Series

Telescoping Series

Requirements for Telescoping Series Test:

• The argument of the series is a rational function with

  • constant in numerator;

  • factorable quadratic in the denominator

• The Method of Partial Fractions can be used to rewrite rational expression as a sum of two partial fractions whose denominators have linear terms.

Definition: Telescoping Series Test

Given the series, ${\displaystyle \sum _{n=j}^{\mathrm{\infty }}\frac{d}{a{n}^2+bn+c}}$$\displaystyle\sum_{n=j}^{\infty}{\frac{d}{an^2+bn+c}}$, where $a,b,c,d\in \mathbb{R}$$a,b,c,d\in\mathbb{R}$,

then the Method of Partial Fractions can be used to separate

${\displaystyle \sum _{n=j}^{\mathrm{\infty }}\frac{d}{a{n}^2+bn+c}=\sum _{n=j}^{\mathrm{\infty }}({\displaystyle \frac{A}{sn+t}}+{\displaystyle \frac{B}{un+v}})}$$\displaystyle\sum_{n=j}^{\infty}{\frac{d}{an^2+bn+c}}=\sum_{n=j}^{\infty}\left(\dfrac{A}{sn+t}+\dfrac{B}{un+v}\right)$

Write out the first several terms (as many as needed until you notice the pattern) of the series and simplify.

Example 01:

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{3}{n^2+5n}=\sum _{n=1}^{\mathrm{\infty }}[\frac{3/5}{n}-\frac{3/5}{n+5}]}$$\displaystyle\sum_{n=1}^{\infty}{\frac{3}{n^2+5n}} =\sum_{n=1}^{\infty}{\left[\frac{3/5}{n}-\frac{3/5}{n+5}\right]}$

It may be helpful to simplify the series by factoring out ${\displaystyle \frac{3}{5}}$$\dfrac{3}{5}$:

${\displaystyle \Rightarrow \sum _{n=1}^{\mathrm{\infty }}[\frac{3/5}{n}-\frac{3/5}{n+5}]={\displaystyle \frac{3}{5}}\sum _{n=1}^{\mathrm{\infty }}[\frac{1}{n}-\frac{1}{n+5}]}$$\displaystyle\Rightarrow\sum_{n=1}^{\infty}{\left[\frac{3/5}{n}-\frac{3/5}{n+5}\right]}=\dfrac{3}{5}\sum_{n=1}^{\infty}{\left[\frac{1}{n}-\frac{1}{n+5}\right]}$

Write out the first several terms of the series and simplify each fraction in each term.

The first term of the series, when $n=1$$n=1$, is $({\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{6}})$$\left(\dfrac{1}{1}-\dfrac{1}{6}\right)$

The second term of the series, when $n=2$$n=2$, is $({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{7}})$$\left(\dfrac{1}{2}-\dfrac{1}{7}\right)$

The third term of the series, when $n=3$$n=3$, is $({\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{8}})$$\left(\dfrac{1}{3}-\dfrac{1}{8}\right)$

The fourth term of the series, when $n=4$$n=4$, is $({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{9}})$$\left(\dfrac{1}{4}-\dfrac{1}{9}\right)$

The fifth term of the series, when $n=5$$n=5$, is $({\displaystyle \frac{1}{5}}-{\displaystyle \frac{1}{10}})$$\left(\dfrac{1}{5}-\dfrac{1}{10}\right)$

The sixth term of the series, when $n=6$$n=6$, is $({\displaystyle \frac{1}{6}}-{\displaystyle \frac{1}{11}})$$\left(\dfrac{1}{6}-\dfrac{1}{11}\right)$

The seventh term of the series, when $n=7$$n=7$, is $({\displaystyle \frac{1}{7}}-{\displaystyle \frac{1}{12}})$$\left(\dfrac{1}{7}-\dfrac{1}{12}\right)$

etc.

Thus, ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{3}{n^2+5n}={\displaystyle \frac{3}{5}}\sum _{n=1}^{\mathrm{\infty }}[\frac{1}{n}-\frac{1}{n+5}]}$$\displaystyle\sum_{n=1}^{\infty}{\frac{3}{n^2+5n}} =\dfrac{3}{5}\sum_{n=1}^{\infty}{\left[\frac{1}{n}-\frac{1}{n+5}\right]}$

${\displaystyle ={\displaystyle \frac{3}{5}}[(\frac{1}{1}-\frac{1}{6})+(\frac{1}{2}-\frac{1}{7})+(\frac{1}{3}-\frac{1}{8})+(\frac{1}{4}-\frac{1}{9})+(\frac{1}{5}-\frac{1}{10})+(\frac{1}{6}-\frac{1}{11})+(\frac{1}{7}-\frac{1}{12})+\cdots ]}$$\displaystyle =\dfrac{3}{5}\left[\left(\frac{1}{1}-\frac{1}{6}\right)+\left(\frac{1}{2}-\frac{1}{7}\right)+\left(\frac{1}{3}-\frac{1}{8}\right)+\left(\frac{1}{4}-\frac{1}{9}\right)+\left(\frac{1}{5}-\frac{1}{10}\right)+\left(\frac{1}{6}-\frac{1}{11}\right)+\left(\frac{1}{7}-\frac{1}{12}\right)+\cdots\right]$

${\displaystyle ={\displaystyle \frac{3}{5}}[(\frac{1}{1}-\cancel{\frac{1}{6}})+(\frac{1}{2}-\bcancel{\frac{1}{7}})+(\frac{1}{3}-\frac{1}{8})+(\frac{1}{4}-\frac{1}{9})+(\frac{1}{5}-\frac{1}{10})+(\cancel{\frac{1}{6}}-\frac{1}{11})+(\bcancel{\frac{1}{7}}-\frac{1}{12})+\cdots ]}$$\displaystyle =\dfrac{3}{5}\left[\left(\frac{1}{1}-\cancel{\frac{1}{6}}\right)+\left(\frac{1}{2}-\bcancel{\frac{1}{7}}\right)+\left(\frac{1}{3}-\frac{1}{8}\right)+\left(\frac{1}{4}-\frac{1}{9}\right)+\left(\frac{1}{5}-\frac{1}{10}\right)+\left(\cancel{\frac{1}{6}}-\frac{1}{11}\right)+\left(\bcancel{\frac{1}{7}}-\frac{1}{12}\right)+\cdots\right]$

If you write out more terms, you will keep canceling terms leaving the following:

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{3}{n^2+5n}={\displaystyle \frac{3}{5}}[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}]=\frac{137}{100}}$$\displaystyle\sum_{n=1}^{\infty}{\frac{3}{n^2+5n}}=\dfrac{3}{5}\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right]=\frac{137}{100}$

Check Your Work

Use Technology to Verify Separation of a Rational Expression into Sum of Partial Fractions

NOTE: The use of technology in this activity is to verify, i.e., check, your partial fractions. You must show work on assessments to receive credit for all Calculus II work.

Mathematica

1
(* Example from Example 01 *)
2
Apart[3/(n^2+5n)]

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. Apart[ ] has one argument:

   1. The rational expression.

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

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

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

   3. Click on local

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

4. You may need parens, $($$(\,$ and $)$$\,)$, to group multiple terms in the numerator and denominator or to group $-1$$-1$ when raised to $n$$n$.

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

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

Investigation 12

For each of the following series,

• Determine if the series is a telescoping series. Explain.

• If the series is a telescoping series, then separate $a_n$$a_n$ into partial fractions.

• Evaluate the first several terms and simplify.

• If possible, determine the sum $S$$S$ of the telescoping series.

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

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

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

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

Homework

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

• Section 11.2: # 43, 45, 47.

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Created: Thursday, 12 November 2020 6:42 EDT Last Modified: Wednesday, 27 October 2021 - 07:00 (EDT)