/*! 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 31 Prove that if \(\sum_{n=1}^{+\in... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 31

Prove that if \(\sum_{n=1}^{+\infty} u_{n}\) is absolutely convergent, then \(\sum_{n=1}^{+\infty} u_{n}{ }^{2}\) is convergent.

Short Answer

Expert verified
Since \( |u_n^2| \leq |u_n| \) and \( \sum_{n=1}^{\infty} |u_n| \) converges, \( \sum_{n=1}^{\infty} u_n^2 \) converges by comparison.

Step by step solution

01

Understand the definition of absolute convergence

A series \( \sum_{n=1}^{\infty} u_n \) is said to be absolutely convergent if the series \( \sum_{n=1}^{\infty} |u_n| \) converges.
02

Given series is absolutely convergent

Since we know \( \sum_{n=1}^{\infty} u_n \) is absolutely convergent, \( \sum_{n=1}^{\infty} |u_n| \) converges.
03

Apply comparison test for convergence

Use the comparison test: If \( |v_n| \leq |u_n| \) and \( \sum |u_n| \) converges, then \( \sum |v_n| \) also converges.
04

Compare \( u_n^2\) to \( u_n\)

Notice that for large n, \( |u_n| \geq |u_n^2| \). Since \( \sum_{n=1}^{\infty} |u_n| \) converges, by the comparison test \( \sum_{n=1}^{\infty} u_n^2 \) must also converge.
05

Conclude the proof

Since \( u_n^2 \leq |u_n| \) and \( \sum_{n=1}^{\infty} |u_n| \) converges, the series \( \sum_{n=1}^{\infty} u_n^2 \) also converges by comparison.

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 fundamental tool in determining the convergence of series. It helps us compare a complex series to a simpler one we know more about. For instance, if you have a series \( \sum_{n=1}^{\+\infty} a_{n} \) and another series \( \sum_{n=1}^{\+\infty} b_{n} \), you can use the comparison test to determine convergence or divergence of the second series based on the first.

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

Compute the value of the given integral, accurate to four decimal places, by using series.\(\int_{0}^{1} h(x) d x\), where \(h(x)= \begin{cases}\frac{\sinh x}{x} & \text { if } x \neq 0 \\ 1 & \text { if } x=0\end{cases}\)

Find a power-series representation for the given function at the number \(a\) and determine its radius of convergence.\(f(x)=\sqrt{x} ; a=4\)

Compute the value of the given integral, accurate to four decimal places, by using series.\(\int_{0}^{1} e^{-x^{2}} d x\)

Use a binomial series to find the Maclaurin series for the given function. Determine the radius of convergence of the resulting series.\(f(x)=\frac{x^{2}}{\sqrt{1+x}}\)

Find the sum of the given infinite series, accurate to three decimal places.\(\sum_{n=1}^{+\infty}(-1)^{n} \frac{1}{(2 n+1)^{3}}\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

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.