/*! 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 In Exercises \(1-36\) , (a) find... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises \(1-36\) , (a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely, (c) conditionally? $$ \sum_{n=1}^{\infty} \sqrt[n]{n}(2 x+5)^{n} $$

Short Answer

Expert verified
Radius of convergence: 1; Interval: \((-3, -2)\); No absolute or conditional convergence.

Step by step solution

01

Identify the Given Power Series

The series given is \( \sum_{n=1}^{\infty} \sqrt[n]{n}(2x+5)^{n} \). This is a power series in the form \( \sum a_n (x - c)^n \) where \( a_n = \sqrt[n]{n} \) and \( c = -\frac{5}{2} \).
02

Use the Root Test to Find Radius of Convergence

Apply the root test, which gives the radius of convergence for this type of series. For the root test, consider:\[ \lim_{n \to \infty} \sqrt[n]{|a_n (2x+5)^n|} = \lim_{n \to \infty} \sqrt[n]{|\sqrt[n]{n} (2x+5)^n|} = \lim_{n \to \infty} n^{1/n} |2x+5| \]We know \( n^{1/n} \to 1 \) as \( n \to \infty \). Therefore, the limit simplifies to \( |2x+5| \). For convergence, we need:\[ |2x+5| < 1 \]
03

Solve for the Interval of Convergence

Solve the inequality \( |2x+5| < 1 \) to find the interval of convergence:1. \( -1 < 2x + 5 < 1 \)2. Solve for \( x \): - Subtract 5 from all parts: \( -6 < 2x < -4 \) - Divide all parts by 2: \( -3 < x < -2 \)Thus, the interval of convergence is \((-3, -2)\).
04

Check the Endpoints for Absolute and Conditional Convergence

Now, test the endpoints \( x = -3 \) and \( x = -2 \) for convergence.1. For \( x = -3 \): - The series becomes \( \sum \sqrt[n]{n}(-1)^n \). The absolute value grows as \( n^{1/n} \). Thus, it doesn't converge absolutely or conditionally.2. For \( x = -2 \): - The series becomes \( \sum \sqrt[n]{n}(1)^n \). Similarly, \( n^{1/n} \) prevents it from converging absolutely or conditionally.Hence, there is no conditional or absolute convergence on the endpoints.

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.

Interval of Convergence
When examining a power series, one essential aspect is identifying where the series converges. This is known as the interval of convergence. For the series \( \sum_{n=1}^{\infty} \sqrt[n]{n}(2x+5)^{n} \), the interval of convergence tells us the range of \(x\) values at which the series converges. To determine this interval, we use the root test, which involves taking the \(n\)-th root of the absolute terms of the series and finding the limit as \(n\) approaches infinity. For the given series, we have\[ |2x+5| < 1 \]By solving this inequality, we find that \(-3 < x < -2\), which represents the interval within which the series converges. So, whenever \(x\) lies between \(-3\) and \(-2\) (not including the endpoints), the series will converge. This defines our interval of convergence, giving us a clear picture of where our series behaves nicely.
Absolute Convergence
Absolute convergence of a series occurs when the series of the absolute values of its terms converges. Looking at the endpoints of our interval of convergence, we will test for absolute convergence by replacing \(x\) with the endpoint values. First, consider \(x = -3\), the series becomes \( \sum_{n=1}^{\infty} \sqrt[n]{n}(-1)^n \). Evaluating the behavior of this series, the absolute terms \(\sqrt[n]{n}\) still diverge as \(n\) grows larger. Thus, it fails to converge absolutely at \(x = -3\). Similarly, for \(x = -2\), the series reduces to \( \sum_{n=1}^{\infty} \sqrt[n]{n}(1)^n \). Again, the term \( \sqrt[n]{n} \) diverges, preventing absolute convergence at this endpoint. In this particular case, neither endpoint values lead to convergence of the absolute series, confirming that absolute convergence is not present at the boundaries of our interval.
Conditional Convergence
Conditional convergence is a slightly less restrictive form of convergence. A series is conditionally convergent if the series itself converges, but the series of its absolute values does not. To explore this in the context of our series, we again test the endpoints of the interval of convergence. For \(x = -3\), as observed earlier, the series \( \sum_{n=1}^{\infty} \sqrt[n]{n}(-1)^n \) doesn't converge, neither absolutely nor conditionally. Similarly, for \(x = -2\), the series becomes \( \sum_{n=1}^{\infty} \sqrt[n]{n}(1)^n \). Once more, it doesn't meet the criteria for convergence. Thus, at neither endpoint do we find conditional convergence. In situations like this, where neither kind of convergence is occurring at endpoints, it highlights the specific range where the series maintains regular convergence.

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

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{2^{n}+3^{n}}{3^{n}+4^{n}}\end{equation}

Show that the sum of the first 2\(n\) terms of the series $$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\cdots$$ is the same as the sum of the first \(n\) terms of the series $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\frac{1}{5 \cdot 6}+\cdots$$ Do these series converge? What is the sum of the first \(2 n+1\) terms of the first series? If the series converge, what is their sum?

\begin{equation} \begin{array}{l}{\text { a. Use Taylor's formula with } n=2 \text { to find the quadratic }} \\ {\text { approximation of } f(x)=(1+x)^{k} \text { at } x=0(k \text { a constant) }} \\ {\text { b. If } k=3, \text { for approximately what values of } x \text { in the interval }} \\ {[0,1] \text { will the error in the quadratic approximation be less }} \\ {\text { than } 1 / 100 ?}\end{array} \end{equation}

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{\tan ^{-1} n}{n^{1.1}}\end{equation}

In the alternating harmonic series, suppose the goal is to arrange the terms to get a new series that converges to \(-1 / 2 .\) Start the new arrangement with the first negative term, which is \(-1 / 2 .\) Whenever you have a sum that is less than or equal to \(-1 / 2,\) start introducing positive terms, taken in order, until the new total is greater than \(-1 / 2 .\) Then add negative terms until the total is less than or equal to \(-1 / 2\) again. Continue this process until your partial sums have been above the target at least three times and finish at or below it. If \(s_{n}\) is the sum of the first \(n\) terms of your new series, plot the points \(\left(n, s_{n}\right)\) to illustrate how the sums are behaving.

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.