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Chapter 11: Problem 28

Which of the series 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鈥� convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{n}{n^{2}+1} $$

Short Answer

Expert verified
The series diverges by the limit comparison test with the harmonic series.

Step by step solution

01

Recognize the Type of Series

The given series is \( \sum_{n=1}^{\infty} \frac{n}{n^{2}+1} \). This is an infinite series where each term is a rational function of \( n \). Our first task is to try and understand the series type to identify an appropriate test for convergence.
02

Apply the Limit Comparison Test

Given \( a_n = \frac{n}{n^2+1} \), let's compare it to the series \( b_n = \frac{1}{n} \), which is a known harmonic series that diverges. We compute the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n}{n^2+1} \cdot n = \lim_{n \to \infty} \frac{n^2}{n^2+1} \). This limit evaluates to 1, meaning both series \( a_n \) and \( b_n \) will have the same behavior since the limit is non-zero and finite.
03

Conclude Based on Comparison

The harmonic series \( \sum \frac{1}{n} \) is known to diverge. Since \( \lim_{n \to \infty} \frac{a_n}{b_n} = 1 \) and is finite and non-zero, the series \( \sum_{n=1}^{\infty} \frac{n}{n^{2}+1} \) must also diverge by the limit comparison test.

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Key Concepts

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

Limit Comparison Test
The Limit Comparison Test is a valuable tool in determining the convergence or divergence of an infinite series. It involves comparing a series with another series that is already known to either converge or diverge. Here's how it works:
  • Select a series, say \( b_n \), with known behavior (either convergent or divergent).
  • Calculate the limit: \( \lim_{n \to \infty} \frac{a_n}{b_n} \), where \( a_n \) are the terms of the series you're investigating.
  • If this limit is a non-zero finite number, the series \( a_n \) behaves like \( b_n \).

In the exercise, the series \( \sum_{n=1}^{\infty} \frac{n}{n^2+1} \) was compared to the harmonic series \( \sum \frac{1}{n} \), which is known to diverge. The limit evaluated to 1, hence the series diverges similarly.
Harmonic Series
The harmonic series is a fundamental concept in mathematical analysis. It is expressed as \( \sum_{n=1}^{\infty} \frac{1}{n} \). One notable characteristic of the harmonic series is that it diverges, meaning as more terms are added, the sum grows without bound.

Despite each individual term decreasing towards zero, the entire series does not settle to a finite sum. This makes the harmonic series an important benchmark for comparison when using tests like the Limit Comparison Test. In our exercise, it served as the known divergent benchmark for comparing with the series \( \sum_{n=1}^{\infty} \frac{n}{n^2+1} \).
Divergence
Divergence refers to the behavior of a series that doesn't approach a finite limit as more terms are added. In simpler terms, the sum keeps increasing or decreasing without reaching a specific value. There are various ways to prove divergence, with the harmonic series being a classic example.

In the context of the given series \( \sum_{n=1}^{\infty} \frac{n}{n^2+1} \), we determined divergence using the Limit Comparison Test against the harmonic series. Both series behave the same at infinity, a key insight paving the way to conclude divergence.
Rational Function
A rational function is a ratio of two polynomials. In our exercise, terms in the series \( \sum_{n=1}^{\infty} \frac{n}{n^2+1} \) are taken from a rational function: \( \frac{n}{n^2+1} \).

Rational functions are significant in analyzing series because they often arise in real-world problems and can be decomposed or simplified into other functions, such as geometric or harmonic ones, to check convergence. In this problem, the form of the rational function helped us identify a suitable comparison series for the Limit Comparison Test.
Infinite Series
An infinite series is a sum of an infinite sequence of terms. These series can either converge to a specific number or diverge, growing indefinitely. Analyzing infinite series is crucial in mathematics, providing insights into diverse fields like engineering and physics.

Various tests, such as the Limit Comparison Test, are utilized to explore these series. Understanding the terms and the series' behavior at large \( n \) helps decide whether it converges (i.e., the terms approach zero fast enough) or diverges. In our step-by-step exercise, \( \sum_{n=1}^{\infty} \frac{n}{n^2+1} \) was investigated to establish its divergence using mathematical comparison.

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Most popular questions from this chapter

According to the Alternating Series Estimation Theorem, how many terms of the Taylor series for tan \(^{-1} 1\) would you have to add to be sure of finding \(\pi / 4\) with an error of magnitude less than \(10^{-3} ?\) Give reasons for your answer.

Estimate the error if \(\cos \sqrt{t}\) is approximated by \(1-\frac{t}{2}+\frac{t^{2}}{4 !}-\frac{t^{3}}{6 !}\) in the integral \(\int_{0}^{1} \cos \sqrt{t} d t\)

According to a front-page article in the December \(15,1992,\) issue of the Wall Street Journal, Ford Motor Company used about 7\(\frac{1}{4}\) hours of labor to produce stampings for the average vehicle, down from an estimated 15 hours in \(1980 .\) The Japanese needed only about 3\(\frac{1}{2}\) hours. Ford's improvement since 1980 represents an average decrease of 6\(\%\) per year. If that rate continues, then \(n\) years from 1992 Ford will use about $$ S_{n}=7.25(0.94)^{n} $$ hours of labor to produce stampings for the average vehicle. Assuming that the Japanese continue to spend 3\(\frac{1}{2}\) hours per vehicle, how many more years will it take Ford to catch up? Find out two ways: a. Find the first term of the sequence \(\left\\{S_{n}\right\\}\) that is less than or equal to \(3.5 .\) b. Graph \(f(x)=7.25(0.94)^{x}\) and use Trace to find where the graph crosses the line \(y=3.5 .\)

Which of the series 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鈥� convergence or divergence.) $$ \sum_{n=3}^{\infty} \frac{(1 / n)}{(\ln n) \sqrt{\ln ^{2} n-1}} $$

Show that the Taylor series for \(f(x)=\tan ^{-1} x\) diverges for \(|x|>1\)
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