/*! 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 34 Show that \(\sum_{n=1}^{\infty} ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that \(\sum_{n=1}^{\infty} \frac{1}{\left(\ln \left(2^{n}\right)\right)^{2}}\) converges..

Short Answer

Expert verified
The series converges by comparison to a convergent p-series.

Step by step solution

01

Simplify the Series

First, simplify the term inside the series. We have \( a_n = \frac{1}{(\ln(2^n))^2} \). Using the logarithmic identity \( \ln(a^b) = b\ln(a) \), we find \( \ln(2^n) = n\ln(2) \). Therefore, \( a_n = \frac{1}{(n\ln(2))^2} = \frac{1}{n^2 (\ln(2))^2} \).
02

Recognize as a p-series Comparison

The series \( \sum_{n=1}^{\infty} \frac{1}{n^2 (\ln(2))^2} \) can be viewed in form \( \frac{1}{n^2} \) times the constant \( (\ln(2))^{-2} \). We can compare it to the known convergent \( p \)-series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), which converges since \( p = 2 > 1 \).
03

Test for Convergence Using Comparison Test

Since each term \( \frac{1}{n^2 (\ln(2))^2} \leq \frac{1}{n^2} \) and \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a convergent series, we apply the comparison test. This confirms that \( \sum_{n=1}^{\infty} \frac{1}{n^2 (\ln(2))^2} \) 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 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 useful method in determining the convergence of series in mathematics. It essentially allows us to infer the behavior of a more complicated series by comparing it to a simpler one with well-known convergence properties. Here's how it works:
  • You take a series \(\sum_{n=1}^{\infty} a_n\) and compare it to another series \(\sum_{n=1}^{\infty} b_n\), which is simpler to analyze.
  • If you find that each term of the first series is less than or equal to the corresponding term of a known convergent series (i.e., \(a_n \leq b_n\) for all n), then the series \(\sum_{n=1}^{\infty} a_n\) also converges.
By establishing this comparison, many complicated series can be understood in terms of their simpler counterparts, unlocking a deeper understanding of their behavior.
p-series
A p-series is a specific type of series that takes the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\), where \(p\) is a positive constant. This is a fundamental concept because the convergence of a p-series is determined by the value of \(p\).
  • If \(p > 1\), the series converges.
  • If \(p \leq 1\), the series diverges.
The p-series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) is a classic example of a convergent series, as \(p = 2\) is greater than 1. This property comes in handy in mathematical analysis because it provides a baseline to compare other series against, especially using the Comparison Test. In the context of the original exercise, recognizing the form allows us to assert convergence through a comparative analysis against known convergent behavior.
Logarithmic Identities
Logarithmic identities are key tools in simplifying and transforming expressions, especially in calculus and algebra. One basic identity is \(\ln(a^b) = b \ln(a)\), which simplifies expressions where a logarithm is applied to a power.In our exercise, this identity helps to rewrite \(\ln(2^n)\) as \(n \ln(2)\), effectively manipulating the logarithmic term into a more manageable form. This simplification reveals the true nature of the series under investigation by demonstrating its relationship with a p-series.Understanding these identities allows students to deconstruct complex expressions into recognizable forms, bridging the gap between intimidating expressions and familiar series forms. It's a powerful skill for managing complex equations and enhancing overall comprehension of series behavior.

One App. One Place for Learning.

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

Get started for free

Study anywhere. Anytime. Across all devices.