Alternating Series

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

• Section 11.5 – Alternating Series

Expected Educational Results

• Objective 22–01: I can analytically (i.e., not graphically and not numerically) verify all conditions needed to use the Alternating Series Test for a given series.

• Objective 22–02: I can use the Alternating Series Test to determine if a series converges or diverge.

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.

Alternating Series

Definition: Alternating Series

An Alternating Series is a series whose terms are alternatively positive and negative, i.e., ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}[(-1{)}^n{a}_n]}$$\displaystyle\sum_{n=0}^{\infty}{\left[(-1)^n a_n\right]}$ or ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}[(-1{)}^{n+1}{a}_n]}$$\displaystyle\sum_{n=0}^{\infty}{\left[(-1)^{n+1} a_n\right]}$

Theorem: Alternating Series Estimation Theorem

Let $S$$S$ be the total sum of the convergent Alternating Series, ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}(-1{)}^n{a}_n}$$\displaystyle\sum_{n=0}^{\infty}{(-1)^n a_n}$. Let $S_n$$S_n$ be the the partial sum of the first $n$$n$ terms. Let $R_n$$R_n$ be the remainder that is defined as the difference between $S$$S$ and $S_n$$S_n$. Then the absolute error is $\left|R_n\right|\le |S-S_n|\le \left|a_{n+1}\right|$$\left|R_n\right|\leq\left|S-S_n\right|\leq \left|a_{n+1}\right|$.

Example 01:

1. Find the partial sum (to six ($6$$6$) decimal places), $S_{11}$$S_{11}$ for ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{1}{n^3}}$$\displaystyle \sum_{n=1}^{\infty}{(-1)^{n+1}\frac{1}{n^3}}$.

2. Find the maximum error of $S_{11}$$S_{11}$.

Solution:

1. $S_{10}=1-0.125+0.037037-0.015625+0.008-0.004630$$S_{10}=1-0.125+0.037037-0.015625+0.008-0.004630$ $+0.002915-0.001953+0.001372-0.001+0.000751=0.901868$$+0.002915-0.001953+0.001372-0.001+0.000751=0.901868$

2. The maximum error by the Alternating Series Estimation Theorem is $\left|R_n\right|\le \left|a_{11+1}\right|=\left|a_{12}\right|=|-0.000578|=0.000578$$\left|R_n\right|\leq\left|a_{11+1}\right|=\left|a_{12}\right|=\left|-0.000578\right|=0.000578$

Investigation 02

Find $S_6$$S_{6}$, to six ($6$$6$) decimal places. Find the maximum error of $S_6$$S_6$.

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

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

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

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

Example 02:

Find the number of terms of ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{1}{n^3}}$$\displaystyle \sum_{n=1}^{\infty}{(-1)^{n+1}\frac{1}{n^3}}$ are needed in the partial sum, $S_n$$S_{n}$, so that the maximum error is $\left|R_n\right|=0.0001$$\left|R_n\right|=0.0001$.

Solution:

By the Alternating Series Estimation Theorem, $\left|R_n\right|\le \left|a_{n+1}\right|=\left|\frac{1}{(n+1{)}^3}\right|=0.0001$$\left|R_n\right|\leq\left|a_{n+1}\right|=\left|\frac{1}{(n+1)^3}\right|=0.0001$

${\displaystyle \Rightarrow \frac{1}{(n+1{)}^3}=0.0001\Rightarrow (n+1{)}^3=\frac{1}{0.0001}=10000}$$\displaystyle\Rightarrow \frac{1}{(n+1)^3}=0.0001\Rightarrow (n+1)^3=\frac{1}{0.0001}=10\,000$

${\displaystyle \Rightarrow n+1=\sqrt[3]{10000}=21.5443\Rightarrow n=20.5443\Rightarrow n=21}$$\displaystyle\Rightarrow n+1=\sqrt[3]{10\,000}=21.5443\Rightarrow n=20.5443\Rightarrow n=21$

Note that $n$$n$ is rounded up.

Investigation 03

Find the number of terms of each of the following series that are needed in the partial sum, $S_n$$S_{n}$, so that the maximum error is $\left|R_n\right|=0.0001$$\left|R_n\right|=0.0001$.

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

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

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

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

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Created: Monday, 20 June 2022 - 21:25 (EDT)