/*! 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 109 Show that if the series \(a_{1... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 109

Show that if the series \(a_{1}+a_{2}+a_{3}+\cdots+a_{n}+\cdots\) converges, then the series \(a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n}+\cdots\) converges also.

Short Answer

Expert verified
The series \(a_{1}+\frac{a_{2}}{2}+...+\frac{a_{n}}{n}\) converges by using the properties of a convergent series and the Cauchy's convergence test.

Step by step solution

01

Write Down the Cauchy's Condition for Convergence

For a series \(a_{1}+a_{2}+...+a_{n}+...\) to converge, it must satisfy the following condition, according to the Cauchy's Convergence Test: \[\forall \epsilon > 0, \exists N \in \mathbb{N} : |s_{m} - s_{n}| < \epsilon, \forall m, n > N\] where \(s_{n}\) is the partial sum of the series, calculated as the sum of the first \(n\) terms.
02

Define the Two Series

Define the two series as following: \(S_n = a_{1}+a_{2}+...+a_{n}\) and \(T_n = a_{1}+\frac{a_{2}}{2}+...+\frac{a_{n}}{n}\). We know that the series \(S_n\) is convergent.
03

Use Cauchy's Condition for the Second Series

Try to derive the Cauchy's condition for the second series, \(T_m - T_n\). Since \(S_n\) is a convergent series, that means for any positive \( \epsilon\), there exist a positive integer \( N\) such that \(|S_m - S_n| < \epsilon\) for all \( m, n > N\). Since each term in the series \(T_n\) is less than or equal to each term \(S_n\), so \(|T_m - T_n| \leq |S_m - S_n|\). This leads to the conclusion that \(|T_m - T_n| < \epsilon\) for all \(m, n > N\).
04

Conclude the Second Series Converges

The result of step 3 fulfills the condition of Cauchy's convergence test, confirming that the series \(T_n = a_{1}+\frac{a_{2}}{2}+...+\frac{a_{n}}{n}\) indeed 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!

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

Probability In Exercises 73 and \(74,\) approximate the normal probability with an error of less than 0.0001 , where the probability is given by $$P(a < x < b)=\frac{1}{\sqrt{2 \pi}} \int_{a}^{b} e^{-x^{2} / 2} d x$$ $$ P(1 < x < 2) $$

Use the Ratio Test or the Root Test to determine the convergence or divergence of the series. \(\frac{1}{(\ln 3)^{3}}+\frac{1}{(\ln 4)^{4}}+\frac{1}{(\ln 5)^{5}}+\frac{1}{(\ln 6)^{6}}+\cdots\)

Finding a Taylor Series In Exercises \(1-12,\) use the definition of Taylor series to find the Taylor series, centered at \(c,\) for the function. $$ f(x)=\sin x, \quad c=\frac{\pi}{4} $$

Verifying a Formula In Exercises 45 and \(46,\) use a power series and the fact that \(i^{2}=-1\) to verify the formula. $$ g(x)=\frac{1}{2 i}\left(e^{i x}-e^{-i x}\right)=\sin x $$

Approximating an Integral In Exercises \(63-70\) , use a power series to approximate the value of the integral with an error of less than \(0.0001 .\) (In Exercises 65 and \(67,\) assume that the integrand is defined as 1 when \(x=0 .\) $$ \int_{0}^{1} e^{-x^{3}} d x $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

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.