/*! 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 20 Establish convergence or diverge... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 20

Establish convergence or divergence by a comparison test. $$ \sum \frac{1}{e^{n}-n^{e}} $$

Short Answer

Expert verified
The series \( \sum \frac{1}{e^n - n^e} \) converges by comparison with the geometric series \( \sum \frac{1}{e^n} \).

Step by step solution

01

Recognize the Series to Compare

To use the comparison test, we need to find a series that is either known to be convergent or divergent. In this case, consider the series \( \sum \frac{1}{e^n} \), which is a geometric series with a common ratio \( r = \frac{1}{e} < 1 \), and is thus convergent.
02

Formulate the Comparison

We observe that \( e^n - n^e > e^n - e^e \) for sufficiently large \( n \). Hence, \( \frac{1}{e^n - n^e} < \frac{1}{e^n - e^e} \). For large \( n \), \( e^e \) is a constant, so this behaves like \( \frac{1}{e^n} \).
03

Apply the Comparison Test

Since \( \sum \frac{1}{e^n} \) converges and \( \frac{1}{e^n - n^e} < \frac{1}{e^n} \) for sufficiently large \( n \), by the direct comparison test, the series \( \sum \frac{1}{e^n - n^e} \) also converges.

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 91影视!

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 when determining the convergence or divergence of series. It's particularly useful when you can relate a complex series to a simpler one that you already understand. Here's how the comparison test works:
  • If you have two series, \(\sum a_n\) and \(\sum b_n\), where \(0 \leq a_n \leq b_n\) for all \(n\) beyond a certain point, and if the larger series \(\sum b_n\) is convergent, so is \(\sum a_n\).
  • If \(a_n \geq b_n \geq 0\) and \(\sum b_n\) is divergent, then \(\sum a_n\) is also divergent.

The trick is to find a "comparison series" that's easier to analyze. For the given problem, the series \(\sum \frac{1}{e^n}\) served as the comparison series due to its known convergence properties. This allows us to conclude that the original series is convergent too.
Geometric Series
A geometric series is one of the most straightforward types of series to understand and test for convergence. It takes the form \(\sum ar^n\), where \(a\) is a constant and \(r\) is the common ratio.
  • The series converges if the absolute value of the common ratio \(|r|\) is less than 1.
  • The sum of the infinite geometric series is \(\frac{a}{1-r}\) when \(|r| < 1\).

In the example given in the problem, the series \(\sum \frac{1}{e^n}\) simplifies to a geometric series with a common ratio \(r = \frac{1}{e}\), which is less than 1. This means that the geometric series converges, and thus provides a foundation for using it as a comparison for the more complex series \(\sum \frac{1}{e^n - n^e}\). Geometric series often serve as excellent benchmarks in comparison tests due to their clear convergence criteria.
Direct Comparison Test
The direct comparison test is a specific type of comparison test used to establish convergence by directly comparing two series.Here's how you apply it:
  • Find a series \(\sum b_n\) that you know converges or diverges.
  • Make sure \(0 \leq a_n \leq b_n\) holds for all large \(n\).
  • Using the relationship, deduce convergence or divergence of \(\sum a_n\).

In the exercise provided, we considered the series \(\sum \frac{1}{e^n}\), which is convergent. The series \(\sum \frac{1}{e^n - n^e}\) was shown to be less than a multiple of this known convergent geometric series for large \(n\). This direct comparison, showing \(\frac{1}{e^n - n^e} < \frac{1}{e^n}\), allows us to conclude using the direct comparison test that the more complex series also converges.

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

Find the first three terms of the Taylor series around \(x=0\). $$ f(x)=\tan ^{-1} x $$

If \(f(x)\) is an even function, so \(f(-x)=f(x),\) what can you say about its Taylor coefficients in \(f=\Sigma a_{n} x^{n} ?\)

Somehow find the first six nonzero terms for \(f=(1-x)\\}\) \(\left(1-x^{3}\right)\).

Find the first three terms of the Taylor series around \(x=0\). $$ f(x)=\tanh ^{-1} x $$

Solve difierential equations by series. From the equation \(d y / d x=y-2\) find all the derivatives of \(y\) at \(x=0\) starting from \(y(0)=1\). Construct the infinite series for \(y\), identify it as a known function, and verify that the function satisfies \(y^{\prime}=y-2\).
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation
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.