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Chapter 10: Problem 24

Which of the series in Exercises \(11-40\) converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{1}{2 n-1} $$

Short Answer

Expert verified
The series \( \sum_{n=1}^{\infty} \frac{1}{2n-1} \) diverges.

Step by step solution

01

Identify the Series

The given series is \( \sum_{n=1}^{\infty} \frac{1}{2n-1} \). This is an infinite series, where each term is \( \frac{1}{2n-1} \).
02

Determine the Type of Series

Recognize that this series is similar to the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \), which is a known divergent series. The terms \( \frac{1}{2n-1} \) are positive and decrease to zero as \( n \to \infty \).
03

Apply the Comparison Test

Compare \( \sum_{n=1}^{\infty} \frac{1}{2n-1} \) with \( \sum_{n=1}^{\infty} \frac{1}{n} \). Since \( \frac{1}{2n-1} > \frac{1}{2n} \) for all \( n \), and the series \( \sum_{n=1}^{\infty} \frac{1}{n} \) diverges, the Comparison Test implies that \( \sum_{n=1}^{\infty} \frac{1}{2n-1} \) also diverges.
04

Conclude the Divergence

Because \( \sum_{n=1}^{\infty} \frac{1}{n} \) diverges and each term of \( \frac{1}{2n-1} \) is greater than \( \frac{1}{2n} \), and since \( \sum_{n=1}^{\infty} \frac{1}{2n} \) also diverges, the series \( \sum_{n=1}^{\infty} \frac{1}{2n-1} \) also diverges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Infinite Series
An infinite series is a summation of an infinite sequence of terms. Rather than having a definitive end, it continues indefinitely. In algebraic notation, we express it as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the terms of the series. In this expression, \( n \) is the index that tells us the position of a term in the sequence. The series examines the behavior of these numbers as \( n \) approaches infinity.
  • **Infinite Series and Convergence**: When we discuss convergence, we're interested in whether the sum of these infinite terms approaches a finite limit.
  • **Infinite Series and Divergence**: Conversely, a series diverges if it doesn't sum to a finite limit.
  • **Key Examples**: Classic examples of infinite series include the geometric series and the harmonic series.
Understanding the behavior of infinite series is vital for determining convergence or divergence, which is often crucial in calculus and mathematical analysis.
Comparison Test
The Comparison Test is a powerful method for determining the convergence or divergence of an infinite series. It involves comparing the terms of a given series with another series whose convergence behavior is already known.
To employ the Comparison Test, follow these steps:
  • Identify a series \( \sum_{n=1}^{\infty} a_n \) and a known series \( \sum_{n=1}^{\infty} b_n \) such that \( a_n \leq b_n \) for all \( n \).
  • If the known series \( \sum_{n=1}^{\infty} b_n \) converges, then the target series \( \sum_{n=1}^{\infty} a_n \) converges as well.
  • Conversely, if \( a_n \geq b_n \) for all \( n \) and \( \sum_{n=1}^{\infty} b_n \) diverges, then \( \sum_{n=1}^{\infty} a_n \) also diverges.
By leveraging the behavior of simpler or well-known series, the Comparison Test allows a broader understanding of more complex series. It is especially useful when dealing with series that can be directly related to classical examples.
Harmonic Series
The harmonic series is one of the most famous examples of a divergent series. It is expressed as \( \sum_{n=1}^{\infty} \frac{1}{n} \). This series consists of summing the reciprocals of all natural numbers.
Let's explore why the harmonic series diverges:
  • **Terms**: The terms of the harmonic series, \( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots \), become smaller but never reach zero.
  • **Behavior**: As the series progresses, even though each individual term decreases, together they accumulate to a sum that grows indefinitely.
  • **Grafual Growth**: This slow, continual growth without bound means the series doesn't approach a finite value; hence, it diverges.
Understanding the divergence of the harmonic series is important as it provides a benchmark for recognizing and comparing the behavior of other series.
Divergent Series
A divergent series is an infinite series with a sum that does not settle to a finite limit. Essential when evaluating series, identifying divergence helps us know when a series doesn't stabilize to a number.
Characteristics of a Divergent Series include:
  • **Non-Convergence**: Unlike convergent series, divergent series go on growing or shrinking without limit.
  • **Examples**: Common examples are the harmonic series and geometric series with certain conditions (where ratio \( |r| \geq 1 \)).
  • **Indicators**: Tests like the Comparison Test often reveal divergence when comparison with a well-known divergent series exists.
Divergent series appear frequently in mathematical and scientific computations. Understanding them allows us to make assumptions about the behavior of complex systems in calculus and real-world models.

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Most popular questions from this chapter

Computer Explorations Taylor's formula with \(n=1\) and \(a=0\) gives the linearization of a function at \(x=0 .\) With \(n=2\) and \(n=3\) we obtain the standard quadratic and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions: \begin{equation} \begin{array}{l}{\text { a. For what values of } x \text { can the function be replaced by each }} \\ {\text { approximation with an error less than } 10^{-2} \text { ? }} \\ {\text { b. What is the maximum error we could expect if we replace the }} \\ {\text { function by each approximation over the specified interval? }}\end{array} \end{equation} Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals in Exercises \(53-58 .\) \begin{equation} \begin{array}{l}{\text { Step } 1 : \text { Plot the function over the specified interval. }} \\ {\text { Step } 2 : \text { Find the Taylor polynomials } P_{1}(x), P_{2}(x), \text { and } P_{3}(x) \text { at }} \\\ {x=0 .}\\\\{\text { Step } 3 : \text { Calculate the }(n+1) \text { st derivative } f^{(n+1)}(c) \text { associated }} \\ {\text { with the remainder term for each Taylor polynomial. }} \\ {\text { Plot the derivative as a function of } c \text { over the specified interval }} \\ {\text { and estimate its maximum absolute value, } M .}\\\\{\text { Step } 4 : \text { Calculate the remainder } R_{n}(x) \text { for each polynomial. }} \\ {\text { Using the estimate } M \text { from Step } 3 \text { in place of } f^{(n+1)}(c), \text { plot }} \\ {R_{n}(x) \text { over the specified interval. Then estimate the values of }} \\ {x \text { that answer question (a). }}\\\\{\text { Step } 5 : \text { Compare your estimated error with the actual error }} \\ {E_{n}(x)=\left|f(x)-P_{n}(x)\right| \text { by plotting } E_{n}(x) \text { over the specified }} \\ {\text { interval. This will help answer question (b). }} \\ {\text { Step } 6 : \text { Graph the function and its three Taylor approximations }} \\ {\text { together. Discuss the graphs in relation to the information }} \\ {\text { discovered in Steps } 4 \text { and } 5 .}\end{array} \end{equation} $$f(x)=e^{-x} \cos 2 x, \quad|x| \leq 1$$

Replace \(x\) by \(-x\) in the Taylor series for \(\ln (1+x)\) to obtain a series for \(\ln (1-x)\) . Then subtract this from the Taylor series for \(\ln (1+x)\) to show that for \(|x|<1\) \begin{equation} \ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right) \end{equation}

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \sin \frac{1}{n}\end{equation}

Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow 0} \frac{\ln \left(1+x^{2}\right)}{1-\cos x} \end{equation}

Series for tan \(^{-1} x\) for \(|x|>1\) Derive the series \begin{equation} \tan ^{-1} x=\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x>1 \end{equation} \begin{equation} \tan ^{-1} x=-\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x<-1 \end{equation} by integrating the series \begin{equation} \frac{1}{1+t^{2}}=\frac{1}{t^{2}} \cdot \frac{1}{1+\left(1 / t^{2}\right)}=\frac{1}{t^{2}}-\frac{1}{t^{4}}+\frac{1}{t^{6}}-\frac{1}{t^{8}}+\cdots \end{equation} in the first case from \(x\) to \(\infty\) and in the second case from \(-\infty\) to \(x\)
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