/*! 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 56 If \(\sum_{n=1}^{\infty} a_{n}\)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 56

If \(\sum_{n=1}^{\infty} a_{n}\) is a convergent series of nonnegative numbers, can anything be said about \(\sum_{n=1}^{\infty}\left(a_{n} / n\right) ?\) Explain.

Short Answer

Expert verified
The series \( \sum_{n=1}^{\infty} \frac{a_{n}}{n} \) converges.

Step by step solution

01

Understanding the Problem

We are given a convergent series \( \sum_{n=1}^{\infty} a_{n} \) with nonnegative terms, and we need to investigate the convergence of the series \( \sum_{n=1}^{\infty} \frac{a_{n}}{n} \). The key here is that the terms \( a_n \) are nonnegative, and we want to see how the denominator \( n \) affects the convergence.
02

Convergence Implication by Comparison

Given that \( \sum_{n=1}^{\infty} a_{n} \) converges, we know that the terms \( a_n \) must get small enough quickly enough for the series to converge. If the terms are nonnegative and a series converges, then for large \( n \), \( a_n \to 0 \). In \( \sum_{n=1}^{\infty} \frac{a_n}{n} \), the terms are smaller than those in \( \sum_{n=1}^{\infty} a_n \) by a factor of \( n \).
03

Application of the Limit Comparison Test

Consider using the Limit Comparison Test with the convergent series \( \sum_{n=1}^{\infty} a_{n} \) and the new series \( \sum_{n=1}^{\infty} \frac{a_{n}}{n} \). For large \( n \), test whether \( \frac{a_{n}/n}{a_{n}} = \frac{1}{n} \to 0 \), indicating that the terms in \( \sum_{n=1}^{\infty} \frac{a_{n}}{n} \) are significantly smaller than those in your reference series \( \sum_{n=1}^{\infty} a_{n} \).
04

Implication of Non-Increasing Sequence

The series \( \sum_{n=1}^{\infty} \frac{a_n}{n} \) involves terms \( \frac{a_n}{n} \) that decrease more rapidly than just \( a_n \) due to the division by \( n \). Thus, \( \sum_{n=1}^{\infty} \frac{a_{n}}{n} \) also converges since the terms become even smaller and the original series already 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.

Limit Comparison Test
The Limit Comparison Test is a valuable tool in determining the convergence of a series. It allows us to compare a given series to another series whose convergence is already known.
  • If you have two series, say \( \sum a_n \) and \( \sum b_n \), the Limit Comparison Test states that if \( \lim_{{n \to \infty}} \frac{a_n}{b_n} = c \), where \( 0 < c < \infty \), then either both series converge or both diverge.
  • This test is particularly useful when evaluating the series \( \sum \frac{a_n}{n} \) when we know \( \sum a_n \) converges. We compare it to \( \sum \frac{1}{n} \), a well-known divergent series.
  • If \( \frac{a_n}{b_n} \) becomes significantly smaller than 1 as \( n \to \infty \), it suggests that \( \sum \frac{a_n}{n} \) converges more easily than \( \sum a_n \).
By using the Limit Comparison Test with \( b_n = a_n \) and considering the limit \( \lim_{{n \to \infty}} \frac{a_n/n}{a_n} = \frac{1}{n} \to 0 \), it indicates that \( \sum \frac{a_n}{n} \) has terms that become negligible quickly.
Convergent Series
A convergent series is a sequence of numbers that sum up to a finite value as the number of terms grows to infinity.
  • A series \( \sum a_n \) converges if the sequence of partial sums \( S_n = a_1 + a_2 + ... + a_n \) approaches a finite limit as \( n \) becomes very large.
  • For a series to be convergent, it is crucial that the terms \( a_n \) decrease towards zero as \( n \) increases.
  • Not every series with terms going to zero is convergent. The famous harmonic series \( \sum \frac{1}{n} \) is a classic example of divergence despite the terms tending closer to zero.
In the context of this exercise, knowing \( \sum a_n \) converges means that the individual terms \( a_n \) shrink to zero sufficiently rapidly. This rapid decrease aids in asserting the behavior of other series derived from \( a_n \), such as \( \sum \frac{a_n}{n} \).
Nonnegative Terms
Nonnegative terms in a series imply that each term is zero or positive. This property simplifies the study of the series' convergence.
  • Nonnegative terms ensure that the series \( \sum a_n \) is monotonically increasing, which simplifies the analysis of convergence.
  • When terms are nonnegative, partial sums consistently grow or remain the same, providing a clearer path to deducing whether the series converges or diverges.
  • For a convergent series with nonnegative terms, the partial sums do not oscillate and always have a definite upper bound.
The nonnegativity in the terms\( a_n \) guarantees that what applies to \( \sum a_n \) in terms of convergence can be manipulated further to understand the effects of dividing each term by \( n \). This manipulation helps prove that \( \sum \frac{a_n}{n} \) will also likely converge under the given conditions.

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

Uniqueness of limits Prove that limits of sequences are unique. That is, show that if \(L_{1}\) and \(L_{2}\) are numbers such that \(a_{n} \rightarrow L_{1}\) and \(a_{n} \rightarrow L_{2},\) then \(L_{1}=L_{2}\).

Show by example that \(\Sigma\left(a_{n} / b_{n}\right)\) may converge to something other than \(A / B\) even when \(A=\Sigma a_{m}, B=\Sigma b_{n} \neq 0,\) and no \(b_{n}\) equals 0

Find the values of \(x\) for which the given geometric series converges. Also, find the sum of the series (as a function of \(x\) ) for those values of \(x .\) $$\sum_{n=0}^{\infty}(-1)^{n} x^{-2 n}$$

Find the first four nonzero terms in the Maclaurin series for the functions. $$e^{\sin x}$$

Use the definition of convergence to prove the given limit. $$\lim _{n \rightarrow \infty}\left(1-\frac{1}{n^{2}}\right)=1$$
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

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.