Alternating Series and Absolute and Conditional Convergence

Alternating Series and the Alternating Series Test

An alternating series is an infinite series in which the terms are alternately positive and negative. An example of an alternating series is equation image indicator $LaTeX: -\frac{1}{2}+\frac{2}{3}-\frac{3}{4}+\frac{4}{5}-\frac{5}{6}+...=\sum^{\infty}_{n=1}\left(-1\right)^n\frac{n}{n+1}$ . Another example of an alternating series is $LaTeX: 1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+...+\frac{\left(-1\right)^{n+1}}{3^{n-1}}+...$ . The Alternating Series Test (also known as Leibniz's Theorem) states that the alternating series equation image indicator $LaTeX: \sum^{\infty}_{n=1}\left(-1\right)^{n+1}a_n=a_1-a_2+a_3-a_4+...$ converges if all three of the following conditions are satisfied:

1. All a_n's are positive.

2. a_n+1 < a_n for all n

3. equation image indicator $LaTeX: \lim_{n \to \infty }a_n=0$

Three applications of the Alternating Series Test are featured in the presentation below.

Two additional examples of using the Alternating Series Test follow.

Alternating Series Test Example 1

Use the alternating series test to determine if the series converges or diverges.

Problem: $LaTeX: 1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+...=\frac{\left(-1\right)^{n+1}}{3^{n-1}}$ text annotation indicator

Solution: All LaTeX: a_n>0

$LaTeX: a_{n+1}\le a_n$ for all n because $LaTeX: \frac{1}{3^n}<\frac{1}{3^{n-1}}$

$LaTeX: \lim_{n \to \infty }a_n=\lim_{n \to \infty }\frac{1}{3^{n-1}}=0$

Thus, the series converges.

Alternating Series Test Example 2

Use the alternating series test to determine if the series converges or diverges.

text annotation indicator Problem: $LaTeX: \frac{1}{\ln2}-\frac{1}{\ln3}+\frac{1}{\ln4}-\frac{1}{\ln5}+...$

Solution: LaTeX: a_n>0

$LaTeX: a_{n+1}\le a_n$ for all n because $LaTeX: \frac{1}{\ln\left(n+1\right)}<\frac{1}{\ln n}$

$LaTeX: \lim_{n \to \infty }a_n=\lim_{n \to \infty }\frac{1}{\ln n}=0$

Thus, the series converges.

Alternating Series Estimation Theorem

A partial sum s_n of any convergent series can be used to approximate the total sum s; however, without knowing the accuracy of the approximation, it is not very useful. The Alternating Series Estimation Theorem gives a method for estimating the size of the error resulting from using only a partial sum of the series. The Alternating Series Estimation Theorem states that if equation image indicator $LaTeX: s=\sum \left(-1\right)^{n+1}a_n$ is the sum of an alternating series that satisfies 0 < a_n+1 < a_n and $LaTeX: \lim_{n \to \infty }a_n=0$ (the three conditions of the Alternating Series Test), then the absolute value of the remainder R_n involved in approximating the sum s by s_n is less than or equal to the numerical value of the first unused term, i.e., |R_n|=| s - s_n | < a_n+1. Because this theorem gives an error bound for the remainder, the theorem is sometimes referred to as the Alternating Series Remainder Theorem. View the examples below to see how the Alternating Series Estimation Theorem is applied.

Alternating Series Estimation Theorem Example 1

Estimate the error using the 8th partial sum of the following series.

Problem: $LaTeX: \sum^\infty _{n=0} \frac{\left(-1\right)^n}{2n}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\frac{1}{64}\Biggr|+\frac{1}{256}-...$

Solution: $LaTeX: 0\le a_{n+1}\le a_n$ for all n and $LaTeX: \lim_{n \to \infty }a_n=0$

- - - Since the series is geometric with $LaTeX: \left|\text{r}\right|<1,\sum$ is given by the formula: $LaTeX: s=\frac{a_1}{1-r}=\frac{a_1}{1-\frac{-1}{2}}=\frac{2}{3}$
    - $LaTeX: s_8=sum\left(seq\left(y_1,x,0,7,1\right)\right)\approx0.664$
    - Error: $LaTeX: \left|R_8\right|=\left|s-s_8\right|=\left|\frac{2}{3}-0.664\right|=0.0026$
    - Since $LaTeX: a_9=\frac{1}{256}\approx0.0039,\:then\:\left|R_8\right|\le a_9$

Alternating Series Estimation Theorem Example 2

Problem: How many terms of the series are needed to find equation image indicator $LaTeX: \sum^{\infty}_{n=1}\left(-1\right)^{n+1}\frac{(-1)^n+1}{n^4}$ with an |error| < 0.001?

Solution: All $LaTeX: 0>0,\:a_{n+1}>a_n$ for all n because $LaTeX: \frac{1}{\left(n+1\right)^4}<\frac{1}{n^4}$ and $LaTeX: \lim_{n \to \infty }a_n=0$ implies that the series converges.

- - - Trial & error: Try $LaTeX: a_5=\frac{1}{5^4}\approx0.0016>0.001$ . Then try $LaTeX: a_6=\frac{1}{6^4}\approx0.00077<0.001$ .
    - Thus $LaTeX: \left|\text{error}\right|=\left|x-x_5\right|<a_6\Longrightarrow n=5$

Absolute and Conditional Convergence

Given any series equation image indicator $LaTeX: \sum a_n$ , it is often helpful to consider the corresponding series whose terms are absolute values of the terms of the original series, i.e., $LaTeX: \sum_{n=1}^{\infty}\left|a_n\right|=\left|a_1\right|+\left|a_2\right|+\left|a_3\right|+...$ . A series $LaTeX: \sum a_n$ is absolutely convergent if the series of absolute values equation image indicator $LaTeX: \sum\left|a_n\right|$ is convergent. A series $LaTeX: \sum a_n$ is conditionally convergent if it is convergent but not absolutely convergent. The next presentation focuses on absolutely and conditionally convergent series.

Use the interactive applet to explore alternating series and absolute convergence by CLICKING HERE.

Absolute and Conditional Convergence Practice

Absolute and Conditional Convergence: Even More Problems!

Complete problems from your textbook and/or online resources as needed to ensure your complete understanding of alternating series and absolute and conditional convergence.

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