/*! 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 64 Show that neither the Ratio Test... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that neither the Ratio Test nor the Root Test provides information about the convergence of $$\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{p}} \quad(p \text { constant })$$

Short Answer

Expert verified
Both Ratio and Root Tests are inconclusive for this series as the limit equals 1.

Step by step solution

01

Understand the Ratio Test

The Ratio Test is applied to a series \(\sum a_n\) and involves the limit \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\). If the limit is less than 1, the series converges absolutely; if greater than 1, it diverges; and if equal to 1, the test is inconclusive.
02

Apply the Ratio Test to the Series

Here, \(a_n = \frac{1}{(\ln n)^p}\). Applying the Ratio Test: \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{1/(\ln(n+1))^p}{1/(\ln n)^p} \right| = \left( \frac{\ln n}{\ln(n+1)} \right)^p \] As \(n \to \infty\), \(\ln(n+1) \approx \ln n\), so the limit becomes 1. The Ratio Test is inconclusive.
03

Understand the Root Test

The Root Test uses the limit \(L = \limsup_{n \to \infty} \sqrt[n]{|a_n|}\). If \(L < 1\), the series converges absolutely; if \(L > 1\), it diverges; if \(L = 1\), the test is inconclusive.
04

Apply the Root Test to the Series

Using \(a_n = \frac{1}{(\ln n)^p}\), apply the Root Test: \[ \limsup_{n \to \infty} \sqrt[n]{\left|a_n\right|} = \limsup_{n \to \infty} \exp\left(-\frac{p}{n} \ln(\ln n)\right) \] Since \(\frac{p}{n}\ln(\ln n)\to 0\) as \(n \to \infty\), the limit is \(\exp(0) = 1\). Thus, the Root Test is also inconclusive.

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.

Ratio Test
The Ratio Test is a powerful tool in determining the convergence of an infinite series. It assesses the limit of the ratio of successive terms in the series. Specifically, if you have a series \(\sum a_n\), the test is carried out by evaluating: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). This expression gives us a limit that helps decide the behavior of the series.
If the calculated limit is less than 1, the series converges absolutely, indicating that it sums to a fixed number. If the limit is greater than 1, the series diverges, implying it grows without bound. However, if the limit equals 1, as is the case with the series in this exercise, the Ratio Test provides no useful conclusion about convergence. This is because the test is inconclusive at this juncture, and additional methods may be required to determine the nature of the series.
Root Test
The Root Test, like the Ratio Test, aims to determine the convergence of an infinite series. It involves taking the _n^\text{th}_ root of the absolute value of the series terms. For a series \( \sum a_n\), you compute: \( L = \limsup_{n \to \infty} \sqrt[n]{|a_n|} \), where \( \limsup \) denotes the limit superior. This test then compares the result with 1.
If \( L < 1 \), the series converges absolutely, indicating a sum total. If \( L > 1 \), it diverges, blowing up to infinity. However, if \( L = 1 \), which happens with some intricate series like the one in this exercise involving logarithms, the Root Test doesn't lead to a definitive answer about the series' convergence. It becomes necessary to apply other methods or tests to assess the series properly.
Series Convergence
Series convergence is a fundamental concept in calculus and analysis, examining whether the sum of infinite terms in a series results in a finite number. Understanding convergence informs if and how series approximate values and functions.
Generally, we're interested in whether a series like \(\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{p}}\) converges, diverges, or whether we need advanced tests like the Ratio or Root Test to delve deeper. When neither the Ratio nor the Root Test leads to a conclusive result, as with this series, it suggests the series requires alternative techniques or tests, like comparison tests or integral tests, to assess its behavior. Being inconclusive means we might be dealing with a borderline case often tied to the intricacies of functions that vary slowly, like logarithms.
Logarithmic Series
Logarithmic series are some of the most fascinating types of series in mathematics, often featuring terms that involve logarithmic functions. These series can sometimes behave in unpredictable ways, influenced by how logarithmic functions themselves grow very slowly.
The series in the example, \( \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{p}} \), highlights the complexity of logarithmic influence, where even slight changes in structure or parameter \( p \) can significantly impact convergence properties. While both the Ratio and Root Tests can often clarify the nature of a series, their inconclusive results with logarithmic series necessitate applying other analytical techniques. This might include the integral test or other forms of comparison to understand whether the series is convergent or divergent.

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

For a sequence \(\left\\{a_{n}\right\\}\) the terms of even index are denoted by \(a_{2 k}\) and the terms of odd index by \(a_{2 k+1} .\) Prove that if \(a_{2 k} \rightarrow L\) and \(a_{2 k+1} \rightarrow L,\) then \(a_{n} \rightarrow L\).

To construct this set, we begin with the closed interval \([0,1] .\) From that interval, remove the middle open interval (1/3, 2/3). leaving the two closed intervals [ 0, 1/3 ] and [2/3, 1 ]. At the second step we remove the open middle third interval from each of those remaining. From \([0.1 / 3]\) we remove the open interval \((1 / 9,2 / 9),\) and from \([2 / 3,1]\) we remove \((7 / 9,8 / 9),\) leaving behind the four closed intervals \([0,1 / 9]\) \([2 / 9,1 / 3],[2 / 3,7 / 9],\) and \([8 / 9,1] .\) At the next step, we remove the middle open third interval from each closed interval left behind, so \((1 / 27,2 / 27)\) is removed from \([0,1 / 9],\) leaving the closed intervals \([0,1 / 27]\) and \([2 / 27,1 / 9] ;(7 / 27,8 / 27)\) is removed from \([2 / 9,1 / 3]\), leaving behind \([2 / 9,7 / 27]\) and \([8 / 27,1 / 3],\) and so forth. We continue this process repeatedly without stopping, at each step removing the open third interval from every closed interval remaining behind from the preceding step. The numbers remaining in the interval \([0,1],\) after all open middle third intervals have been removed, are the points in the Cantor set (named after Georg Cantor, \(1845-1918\) ). The set has some interesting properties. a. The Cantor set contains infinitely many numbers in [0,1] List 12 numbers that belong to the Cantor set. b. Show, by summing an appropriate geometric series, that the total length of all the open middle third intervals that have been removed from [0,1] is equal to 1

Assume that each sequence converges and find its limit. \(a_{1}=-1, \quad a_{n+1}=\frac{a_{n}+6}{a_{n}+2}\)

Assume that each sequence converges and find its limit. \(a_{1}=0, \quad a_{n+1}=\sqrt{8+2 a_{n}}\)

The (second) second derivative test Use the equation $$f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}\left(c_{2}\right)}{2}(x-a)^{2}.$$to establish the following test. Let \(f\) have continuous first and second derivatives and suppose that \(f^{\prime}(a)=0 .\) Then a. \(f\) has a local maximum at \(a\) if \(f^{\prime \prime} \leq 0\) throughout an interval whose interior contains \(a\) b. \(f\) has a local minimum at \(a\) if \(f^{\prime \prime} \geq 0\) throughout an interval whose interior contains \(a\).

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.