/*! 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 8 Verify that the infinite series ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 8

Verify that the infinite series diverges. \(\sum_{n=0}^{\infty}\left(\frac{4}{3}\right)^{n}\)

Short Answer

Expert verified
The infinite series \(\sum_{n=0}^{\infty}\left(\frac{4}{3}\right)^{n}\) diverges.

Step by step solution

01

Understanding Geometric Series

A geometric series is a series with a common ratio between successive terms. It is represented generically by \(\sum_{n=0}^{\infty}ar^{n}\) where \(a\) is first term and \(r\) is the common ratio. A geometric series converges if and only if \(|r| < 1\). If \(|r| > 1\) or \(|r| = 1\), the series diverges.
02

Identifying the common ratio

In this problem, the infinite series given is \(\sum_{n=0}^{\infty}\left(\frac{4}{3}\right)^{n}\). Here, the common ratio \(r\) is \(\frac{4}{3}\).
03

Test the given series for convergence

Now we check if \(|\frac{4}{3}| < 1\). As \(|\frac{4}{3}| = 1.33\) which is greater than 1, according to the convergence condition for geometric series which mentions that \(|r|\) must be less than 1 for series to converge, the series \(\sum_{n=0}^{\infty}\left(\frac{4}{3}\right)^{n}\) diverges.

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.

Geometric Series
A geometric series is a special type of infinite series where each term after the first is obtained by multiplying the previous term by a constant known as the common ratio. The series can be expressed in the form:\[ \sum_{n=0}^{ ext{infinity}} ar^n \]where:
  • \(a\) is the first term of the series,
  • \(r\) is the common ratio.
Understanding the structure and behavior of a geometric series is crucial in determining whether the series will converge (approach a particular value) or diverge (continue indefinitely without settling on any value). By knowing the components of the series鈥攑articularly the common ratio鈥攜ou can discern a lot about the series' behavior.
Convergence Criteria
For geometric series, whether they converge or diverge is determined solely by the value of the common ratio \(r\). These convergence criteria are straightforward:
  • If \(|r| < 1\), the series converges. This means the sum of all terms will tend to a finite limit as more terms are added.
  • If \(|r| \ge 1\), the series diverges. In this case, the terms grow in magnitude, causing the series to expand indefinitely.
For example, in the given infinite series \(\sum_{n=0}^{\infty}\left(\frac{4}{3}\right)^{n}\), we check the absolute value of the common ratio. As the common ratio \(r\) is \(\frac{4}{3}\) and \(|\frac{4}{3}| = 1.33\), which is greater than 1, the series does not meet the convergence criterion. Thus, it will diverge.
Common Ratio
The common ratio \(r\) is a pivotal element in any geometric series. It dictates how each term is generated from the previous term. In our particular series: \[\sum_{n=0}^\infty\left(\frac{4}{3}\right)^n\]The common ratio \(r\) is identified as \(\frac{4}{3}\). Since \(|\frac{4}{3}| = 1.33\), which exceeds 1, it implies that each successive term in the series will be larger than the one before it. This increasing trend directly influences whether the series will converge or diverge.Being able to accurately identify and understand the significance of the common ratio is essential for assessing the behavior of a geometric series. It helps predict if additional terms will lead to a bounding number (convergence) or if they will continuously grow without bound (divergence).

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

Consider the sequence \(\left\\{a_{n}\right\\}=\left\\{\frac{1}{n} \sum_{k=1}^{n} \frac{1}{1+(k / n)}\right\\}\). (a) Write the first five terms of \(\left\\{a_{n}\right\\}\). (b) Show that \(\lim _{n \rightarrow \infty} a_{n}=\ln 2\) by interpreting \(a_{n}\) as a Riemann sum of a definite integral.

Consider the sequence \(\left\\{a_{n}\right\\}\) where \(a_{1}=\sqrt{k}, a_{n+1}=\sqrt{k+a_{n}}\), and \(k>0\). (a) Show that \(\left\\{a_{n}\right\\}\) is increasing and bounded. (b) Prove that \(\lim _{n \rightarrow \infty} a_{n}\) exists. (c) Find \(\lim _{n \rightarrow \infty} a_{n^{\prime}}\)

Prove that the power series \(\sum_{n=0}^{\infty} \frac{(n+p) !}{n !(n+q) !} x^{n}\) has a radius of convergence of \(R=\infty\) if \(p\) and \(q\) are positive integers.

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. Every decimal with a repeating pattern of digits is a rational number.

Prove that if \(\left\\{s_{n}\right\\}\) converges to \(L\) and \(L>0\), then there exists ? number \(N\) such that \(s_{n}>0\) for \(n>N\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision 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.