/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Use the Comparison Test, the Lim... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the Comparison Test, the Limit Comparison Test, or the Integral Test to determine whether the series converges or diverges. $$ \sum_{n=2}^{\infty} \frac{1}{n \sqrt{\ln n}} $$

Short Answer

Expert verified
The series diverges by the Integral Test.

Step by step solution

01

Decide which test to use

Given the series \( \sum_{n=2}^{\infty} \frac{1}{n \sqrt{\ln n}} \), we consider the **Integral Test**. The terms \( a_n = \frac{1}{n \sqrt{\ln n}} \) are positive and decreasing for \( n \geq 2 \). The Integral Test is applicable here.
02

Set up the integral for the Integral Test

Using the function \( f(x) = \frac{1}{x \sqrt{\ln x}} \), we set up the integral \( \int_{2}^{\infty} \frac{1}{x \sqrt{\ln x}} \, dx \). We will evaluate whether this integral converges or diverges.
03

Perform substitution

Let \( u = \ln x \), then \( du = \frac{1}{x} \, dx \). Change the bounds: when \( x = 2 \), \( u = \ln 2 \); when \( x = \infty \), \( u = \infty \). The integral becomes \( \int_{\ln 2}^{\infty} \frac{1}{\sqrt{u}} \, du \).
04

Evaluate the integral

The integral \( \int_{\ln 2}^{\infty} \frac{1}{\sqrt{u}} \, du \) is an improper integral. The antiderivative is \( 2\sqrt{u} \). Evaluating, we find:\[\left[ 2\sqrt{u} \right]_{\ln 2}^{\infty} = \lim_{b \to \infty} \left( 2\sqrt{b} - 2\sqrt{\ln 2} \right) = \infty\].
05

Conclude the test

Since the integral \( \int_{2}^{\infty} \frac{1}{x \sqrt{\ln x}} \, dx \) diverges, by the Integral Test, the series \( \sum_{n=2}^{\infty} \frac{1}{n \sqrt{\ln n}} \) also diverges.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Comparison Test
The Comparison Test is a powerful tool used to determine the convergence or divergence of a series by comparing it to another series whose behavior is already known. To apply this test, we need two series: one being evaluated and another well-known benchmark series. Here's how it works:
  • If the series \( \sum b_n \) converges and the terms are ordered such that \( 0 \leq a_n \leq b_n \) for every \( n \), then \( \sum a_n \) also converges.
  • If the series \( \sum b_n \) diverges and the terms satisfy \( 0 \leq b_n \leq a_n \) for every \( n \), then \( \sum a_n \) also diverges.
By choosing an appropriate series to compare with, the Comparison Test can provide a quick determination of convergence or divergence without direct evaluation. This test works particularly well when dealing with series that resemble known forms, such as geometric or p-series.
Limit Comparison Test
The Limit Comparison Test is a variation of the standard Comparison Test, providing an alternative when direct comparison isn't straightforward. This test uses the limit of the ratio of corresponding terms from the series in question and a similar known series. Here’s how you can apply it:
  • Consider two series: \( \sum a_n \) (the series of interest) and \( \sum b_n \) (a known benchmark series).
  • Calculate the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} \).
  • If \( L \) is a positive finite number (i.e., \( 0 < L < \infty \)), then both series \( \sum a_n \) and \( \sum b_n \) either converge or diverge together.
This test is especially useful when terms from the series of interest closely approximate those of a simpler series. It offers a convenient way to infer the behavior of complicated series based on more manageable ones.
Series Convergence and Divergence
Understanding series convergence and divergence is central when analyzing infinite series. A series is a sum of terms from a sequence and can behave in two major ways: - **Convergence**: The series converges if the sum of its infinite terms approaches a finite number. Convergence indicates that we can add up all the terms endlessly and settle at a specific value. - **Divergence**: A series diverges if its sum doesn't approach any finite limit. It may either increase without bound, oscillate indefinitely, or otherwise fail to settle. Various tests, including the Integral Test, Comparison Test, and Limit Comparison Test, help determine whether a series converges or diverges. Recognizing the behavior of a series is crucial in many areas like mathematical analysis, physics, and engineering.
Improper Integral Evaluation
Evaluating improper integrals is essential when applying the Integral Test for series convergence. An improper integral involves functions that are unbounded or approach infinity over their interval. Here's how to properly evaluate them:- **Setup**: Convert the series into an integral. For a series with terms \( a_n \), express it as a function \( f(x) \) and evaluate \( \int_{a}^{\infty} f(x) \, dx \), where \( a \) is the starting point of your series.- **Substitution**: Sometimes, substitution methods simplify integration. For example, in the exercise we saw, substituting \( u = \ln x \) transformed the integral into a more manageable form.- **Anti-differentiation and Limits**: Find the anti-derivative, then evaluate the limit as the variable approaches infinity or the boundaries causing an improper nature.If the result is finite, the integral converges, indicating the series likewise converges. If it diverges to infinity, the series also diverges. This technique is powerful and precisely illustrated in the solution steps provided.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A rocket is launched from earth. On its journey it consumes one fourth of its fuel during the first 100 miles, one ninth of its initial fuel during the second 100 miles, and in general, \(1 /(n+1)^{2}\) of its initial fucl during the \(n\) th 100 miles. Does the rocket ever use up all its fuel? (Hint: \(\left.\sum_{n=1}^{\infty} 1 / n^{2}=\pi^{2} / 6 .\right)\)

a. Two trains, each traveling 15 miles per hour, approach each other on a straight track. When the trains are 1 mile apart, a bee begins flying back and forth between the trains at 30 miles per hour. Express the distance the bee travels before the trains collide as an infinite series, and find its sum. b. Find a simple solution of the bee problem without using series. (Hint: Determine how long the bee flies.) (It is said that a similar problem was posed to the great twentieth-century mathematician John von Neumann ( \(1903-1957\) ), who solved it almost instantly in his head. When the poser of the problem suggested that by the quickness of his response, he must have solved the problem the simple way, von Neumann replied that he had actually solved the problem by summing a series.)

Find the Taylor series of \(f\) about \(a\), and write out the first four terms of the series. $$ f(x)=\frac{x}{\sqrt{1-x^{2}}} ; a=0 $$

Find the Taylor series of \(f\) about \(a\), and write out the first four terms of the series. $$ f(x)=\frac{1}{\sqrt{1+x}} ; a=0 $$

Find the Taylor series of the given function about \(a\). Use the series already obtained in the text or in previous exercises. $$ f(x)=10^{x} ; a=0 $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.