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

Geometric Series

Definition: Geometric Series

Geometric series have the form ${\displaystyle \sum _{i=0}^{\mathrm{\infty }}c{r}^n}$$\displaystyle\sum_{i=0}^{\infty}{c r^n}$, where $c$$c$ is a constant and $r$$r$ is the common ratio.

Theorem: Convergent Geometric Series

An infinite geometric series

• converges if $\left|r\right|<1$$\left|\,\;r\;\right|<1$ (i.e., $-1<r<1$$-1< r< 1$) and

• diverges for all other $r$$r$ values.

Investigation 04

Let $r=1$$r=1$, find the sum of Partial Sums of the series, ${\displaystyle \sum _{i=0}^{\mathrm{\infty }}{1}^n}$$\displaystyle\sum_{i=0}^{\infty}{1^n}$. What do you notice? Explain.

Investigation 05

Let $r=-1$$r=-1$, find the sum of Partial Sums of the series, ${\displaystyle \sum _{i=0}^{\mathrm{\infty }}(-1{)}^n}$$\displaystyle\sum_{i=0}^{\infty}{(-1)^n}$. What do you notice? Explain.

Definition: Sum of Convergent Geometric Series

Let $a_i$$a_i$ be the first term of a convergent geometric series. Then the sum, $S$$S$, of the convergent geometric series is

$S={\displaystyle \frac{a_i}{1-r}}$$S=\dfrac{a_i}{1-r}$

The derivation of the sum of a convergent geometric series is shown at the bottom of this document.

Investigation 06

For each of the following series, determine

• If the series is a geometric series. Explain.

• If the series is a geometric series, then find the common ratio, $r$$r$. Explain.

• If the series is a convergent or divergent. Explain.

• If the series is a convergent geometric series, then find $S$$S$.

1. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}5{\left(\frac{3}{7}\right)}^{n+1}}$$\displaystyle\sum_{n=0}^{\infty}{5\left(\frac{3}{7}\right)^{n+1}}$

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

3. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}3{\left(\frac{4}{3}\right)}^n}$$\displaystyle\sum_{n=0}^{\infty}{3\left(\frac{4}{3}\right)^n}$

4. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}{(-\frac{1}{5})}^n}$$\displaystyle\sum_{n=0}^{\infty}{\left(-\frac{1}{5}\right)^n}$

5. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\left(\frac{2}{3}\right)}^n}$$\displaystyle\sum_{n=0}^{\infty}{\left(\frac{2}{3}\right)^{n}}$

6. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\left(\frac{5}{2}\right)}^n}$$\displaystyle\sum_{n=0}^{\infty}{\left(\frac{5}{2}\right)^{n}}$

Check Your Work

Use Technology to Verify Sum of Convergent Geometric Series

NOTE: The use of technology in this activity is to verify, i.e., check, your computation of of the sum of a convergent geometric series. You must show work on assessments to receive credit for all Calculus II work.

Mathematica

1
(* Example from Investigation 06-2 *)
2
Sum[(-1/5)^n, {n,0,Infinity}]

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. Sum[ ] has two arguments:

   1. The general $n^{\text{th}}$$n^{\text{th}}$-term, $a_n$$a_n$;

   2. $\{n,a,b\}$$\{n,a,b\}$, where

      1. $n$$n$ is the index (i.e., label for the terms);

      2. $a$$a$ is the initial label;

      3. $b$$b$ is the upper label.

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

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

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

   3. Click on local

   4. Read how to use the Sum[ ] 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.

Power Series

Definition: Power Function

Power functions have the form $x^r$$x^r$, which is a form of a geometric series.

Corollary

An power series, ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}{(ax-b)}^n}$$\displaystyle\sum_{n=0}^{\infty}{\left(ax-b\right)^n}$, converges, by the Geometric Series Test, if $|ax-b|<1$$\left|\,ax-b\,\right|< 1$.

Investigation 07

Find all values for $x$$x$ so that the series converges. Explain.

1. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}3{(2x+1)}^n}$$\displaystyle\sum_{n=0}^{\infty}{3\left(2x+1\right)^{n}}$

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

3. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\left(\frac{2x-1}{3}\right)}^{n+1}}$$\displaystyle\sum_{n=0}^{\infty}{\left(\frac{2x-1}{3}\right)^{n+1}}$

Check Your Work

Use Technology to Solve Polynomial Inequalities

NOTE: The use of technology in this activity is to solve polynomial inequalities. On Assessments, you may use technology to solve polynomial inequalities without showing work.

Mathematica

xxxxxxxxxx
2
1
(* Example from Investigation 05-1 *)
2
Reduce[Abs[2x+1]<1,x,Reals]//Simplify

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. Reduce[ ] has two arguments:

   1. The absolute value inequality to be solved;

   2. $x$$x$, the independent variable;

   3. $\text{Reals}$$\text{Reals}$ to force Mathematica to show only Real-valued solutions.

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

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

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

   3. Click on local

   4. Read how to use the Reduce[ ] 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.

Derivation of the Sum of Convergent Geometric Series

NOTE: You will not be assessed on the derivation of the sum of a convergent geometric series.

Suppose the geometric series, ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}a{r}^n}$$\displaystyle \sum_{n=0}^{\infty}{ar^n}$, converges. In other words, the total sum of the series is a Real number, $S$$S$. Thus,

$S=a+ar+a{r}^2+a{r}^3+a{r}^4+a{r}^5+\cdots$$S=a+ar+ar^2+ar^3+ar^4+ar^5+\cdots$

Multiply all terms in the above equation with $-r$$-r$:

$-rS=-ar-a{r}^2-a{r}^3-a{r}^4-a{r}^5-\cdots$$-rS=-ar-ar^2-ar^3-ar^4-ar^5-\cdots$

Add these two equations

$\begin{array}{c}\phantom{\text{xxlx}}S=a\phantom{\text{x}}+\bcancel{ar}+\cancel{a{r}^2}+\bcancel{a{r}^3}+\cancel{a{r}^4}+\bcancel{a{r}^5}+\cdots \\ +(-rS=\phantom{\text{xx}}\bcancel{-ar}\cancel{-a{r}^2}\bcancel{-a{r}^3}\cancel{-a{r}^4}\bcancel{-a{r}^5}-\cdots )\end{array}$$\phantom{\text{xxlx}} S = a \phantom{\text{x}} \,+ \bcancel{ar}+ \cancel{ar^2} \,+ \bcancel{ar^3} + \cancel{ar^4} + \bcancel{ar^5} + \cdots \\ +\left(-rS = \phantom{\text{xx}}\bcancel{-ar}\cancel{-\,ar^2}\bcancel{-\,ar^3}\cancel{-\,ar^4}\!\bcancel{-\,ar^5}-\cdots\right)$

which results in

$S-rS=a$$S-rS=a$

Factor $S$$S$ from the two terms in the left-side of the above equation:

$\Rightarrow S(1-r)=a\Rightarrow S={\displaystyle \frac{a}{1-r}}$$\Rightarrow S(1-r)=a\Rightarrow S=\dfrac{a}{1-r}$

Solve for $S$$S$:

$\Rightarrow S={\displaystyle \frac{a}{1-r}}{}_{◼}$$\Rightarrow S=\dfrac{a}{1-r}\,\,_{\blacksquare}$

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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, 26 October 2021 - 18:33 (EDT)