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 Application

Rewrite Repeating Decimal as a Fraction

Example 01: Use a series to rewrite $0.3\overline{52}$$0.3\overline{52}$ as a ratio of two integers.

Solution:

$\begin{array}{c}{\displaystyle 0.3\overline{52}=0.35252525252\dots }\\ =0.3+0.052+0.00052+0.0000052+0.000000052+0.00000000052+\cdots \\ ={\displaystyle \frac{3}{10}}+{\displaystyle \frac{52}{1000}}+{\displaystyle \frac{52}{100000}}+{\displaystyle \frac{52}{10000000}}+{\displaystyle \frac{52}{1000000000}}+{\displaystyle \frac{52}{100000000000}}+\cdots \\ ={\displaystyle \frac{3}{10}}+[{\displaystyle \frac{52}{{10}^3}}+{\displaystyle \frac{52}{{10}^5}}+{\displaystyle \frac{52}{{10}^7}}+{\displaystyle \frac{52}{{10}^9}}+{\displaystyle \frac{52}{{10}^{11}}}+\cdots ]\end{array}$$\displaystyle 0.3\overline{52} = 0.35252525252\ldots \\ = 0.3 + 0.052 + 0.00052 + 0.0000052 + 0.000000052 + 0.00000000052 + \cdots \\ = \dfrac{3}{10} + \dfrac{52}{1000} + \dfrac{52}{100000} + \dfrac{52}{10000000} + \dfrac{52}{1000000000} + \dfrac{52}{100000000000} + \cdots \\= \dfrac{3}{10} + \left[\dfrac{52}{10^3} + \dfrac{52}{10^5} + \dfrac{52}{10^7} + \dfrac{52}{10^9} + \dfrac{52}{10^{11}} + \cdots\right]$

The infinite series in the square brackets above is a geometric series with ${\displaystyle a_0=\frac{52}{{10}^3}}$$\displaystyle a_0=\frac{52}{10^3}$ and ${\displaystyle r=\frac{1}{{10}^2}}$$\displaystyle r=\frac{1}{10^2}$.

Since ${\displaystyle \left|\frac{1}{{10}^2}\right|<1}$$\displaystyle \left|\frac{1}{10^2}\right|<1$, the geometric series converges and ${\displaystyle S=\frac{\frac{52}{{10}^3}}{1-\frac{1}{{10}^2}}}$$\displaystyle S=\frac{\frac{52}{10^3}}{1-\frac{1}{10^2}}$

There are many ways to rewrite the sum of the geometric series, here is one way:

${\displaystyle S=\frac{\frac{52}{{10}^3}}{1-\frac{1}{{10}^2}}\left(\frac{{10}^3}{{10}^3}\right)=\frac{52}{{10}^3-10}=\frac{52}{1000-10}=\frac{52}{990}}$$\displaystyle S=\frac{\frac{52}{10^3}}{1-\frac{1}{10^2}}\left(\frac{10^3}{10^3}\right)=\frac{52}{10^3-10}=\frac{52}{1000-10}=\frac{52}{990}$

So, ${\displaystyle 0.35252525252\dots =\frac{3}{10}+\frac{52}{990}=\frac{3}{10}\left(\frac{99}{99}\right)+\frac{52}{990}=\frac{297}{990}+\frac{52}{990}=\frac{349}{990}}$$\displaystyle 0.35252525252\ldots = \frac{3}{10} + \frac{52}{990} = \frac{3}{10}\left(\frac{99}{99}\right) + \frac{52}{990} = \frac{297}{990} + \frac{52}{990} = \frac{349}{990}$

Check: ${\displaystyle \frac{349}{990}}=0.3525252525252525\dots$$\dfrac{349}{990}=0.3525252525252525\ldots$

Example 02: Use a series to rewrite $0.01\overline{917}$$0.01\overline{917}$ as a ratio of two integers.

Solution:

$\begin{array}{c}{\displaystyle 0.01\overline{917}=0.01917917917\dots }\\ =0.01+0.00917+0.00000917+0.00000000917+\cdots \\ ={\displaystyle \frac{1}{100}}+{\displaystyle \frac{917}{100000}}+{\displaystyle \frac{917}{100000000}}+{\displaystyle \frac{917}{100000000000}}+\cdots \\ ={\displaystyle \frac{1}{100}}+[{\displaystyle \frac{917}{{10}^5}}+{\displaystyle \frac{917}{{10}^8}}+{\displaystyle \frac{917}{{10}^{11}}}+\cdots ]\end{array}$$\displaystyle 0.01\overline{917} = 0.01917917917\ldots \\ = 0.01 + 0.00917 + 0.00000917 + 0.00000000917 + \cdots \\ = \dfrac{1}{100} + \dfrac{917}{100000} + \dfrac{917}{100000000} + \dfrac{917}{100000000000} + \cdots \\ = \dfrac{1}{100} + \left[\dfrac{917}{10^5} + \dfrac{917}{10^8} + \dfrac{917}{10^{11}} + \cdots\right]$

The infinite series in the square brackets above is a geometric series with ${\displaystyle a_0=\frac{917}{{10}^5}}$$\displaystyle a_0=\frac{917}{10^5}$ and ${\displaystyle r=\frac{1}{{10}^3}}$$\displaystyle r=\frac{1}{10^3}$.

Since ${\displaystyle \left|\frac{1}{{10}^3}\right|<1}$$\displaystyle \left|\frac{1}{10^3}\right|<1$, the geometric series converges and ${\displaystyle S=\frac{\frac{917}{{10}^5}}{1-\frac{1}{{10}^3}}}$$\displaystyle S=\frac{\frac{917}{10^5}}{1-\frac{1}{10^3}}$

There are many ways to rewrite the sum of the geometric series, here is one way:

${\displaystyle S=\frac{\frac{917}{{10}^5}}{1-\frac{1}{{10}^3}}\left(\frac{{10}^5}{{10}^5}\right)=\frac{917}{{10}^5-{10}^2}=\frac{917}{100000-100}=\frac{917}{99900}}$$\displaystyle S=\frac{\frac{917}{10^5}}{1-\frac{1}{10^3}}\left(\frac{10^5}{10^5}\right)=\frac{917}{10^5-10^2}=\frac{917}{100000-100}=\frac{917}{99900}$

So, ${\displaystyle 0.01917917917\dots =\frac{1}{100}+\frac{917}{99900}=\frac{1}{100}\left(\frac{999}{999}\right)+\frac{917}{99900}=\frac{999}{99900}+\frac{917}{99900}=\frac{1916}{99900}}$$\displaystyle 0.01917917917\ldots = \frac{1}{100} + \frac{917}{99900} = \frac{1}{100}\left(\frac{999}{999}\right) + \frac{917}{99900} = \frac{999}{99900} + \frac{917}{99900} = \frac{1916}{99900}$

Check: ${\displaystyle \frac{1916}{99900}}=0.01917917917917917\dots$$\dfrac{1916}{99900}=0.01917917917917917\ldots$

Investigation 13

Use a series to rewrite the following as a ratio of two integers:

1. $0.7\overline{81}$$0.7\overline{81}$

2. $0.5\overline{183}$$0.5\overline{183}$

3. $0.12\overline{345}$$0.12\overline{345}$

4. $0.74\overline{447}$$0.74\overline{447}$

Homework

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

Section 11.2: # 51, 53, 55.

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Created: Tuesday, 1 December 2020 15:47 EDT Last Modified: Wednesday, 27 October 2021 - 07:07 (EDT)