/*! 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 54 Show that if \(\lim _{n \rightar... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 54

Show that if \(\lim _{n \rightarrow 0}\left|c_{n+1} / c_{n}\right|=L \neq 0\), then the radius of convergence of \(\sum_{n=0}^{\infty} c_{n} x^{n}\) is \(1 / L\).

Short Answer

Expert verified
The radius of convergence is \(1/L\).

Step by step solution

01

Identify the Ratio Test Condition

Recall the ratio test for the convergence of power series. The ratio test states that the radius of convergence, \( R \), for the series \( \sum_{n=0}^{\infty} c_n x^n \) can be found using \( \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = L \). If this limit exists and is equal to \( L \), then the series converges if \( |x| < 1/L \).
02

Consider Series Radius of Convergence

Recognize that if \( \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = L \) and \( L eq 0 \), it provides information about the radius of convergence. Specifically, convergence occurs when \( |x| < 1/L \), so the radius of convergence, \( R \), is \( 1/L \).
03

Apply the Definition for Convergence

Based on the limit, when \( \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = L \), the power series \( \sum_{n=0}^{\infty} c_n x^n \) converges absolutely if the condition \( |x| < 1/L \) holds. Hence, the radius of convergence is determined as \( 1/L \).
04

Conclusion

Therefore, by the ratio test, the radius of convergence of the series \( \sum_{n=0}^{\infty} c_n x^n \) is \( 1/L \) when \( L eq 0 \).

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.

Ratio Test
The Ratio Test is a handy tool used to determine the convergence of a series, particularly a power series. The idea is to examine the limit of the ratio of consecutive terms in the series. If we have a series \( \sum_{n=0}^{\infty} c_n x^n \), the Ratio Test advises looking at \[ \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = L. \]
  • If the limit \( L < 1 \), the series converges absolutely for all \( x \).
  • If \( L > 1 \) or \( L = \infty \), the series diverges.
  • If \( L = 1 \), the test is inconclusive and additional tests might be needed.

This test is particularly useful for determining the radius of convergence, a concept linked closely to power series. In simpler terms, the bolder the step from one term to the next, the stricter our condition for convergence becomes.
Power Series
A power series is a type of infinite series that resembles polynomial expressions but extends indefinitely. It takes the form \[ \sum_{n=0}^{\infty} c_n x^n, \] where \( c_n \) are coefficients, and \( x \) is the variable. These series can be thought of as a generalization of polynomials with infinitely many terms. They are used to represent functions in a more flexible manner.
  • Power series can converge (sum up to a finite value) for certain values of \( x \), which form an interval centered at the origin.
  • The extent of this interval is described by the radius of convergence, \( R \).
  • If \( |x| < R \), the series converges; if \( |x| > R \), it diverges.

The power and beauty of power series lie in their ability to approximate functions around a certain point, offering a precise understanding of function behavior within their radius of convergence.
Absolute Convergence
Absolute convergence is a strong form of convergence concerning series. For a series \( \sum_{n=0}^{\infty} a_n \), absolute convergence implies that the series of absolute values \[ \sum_{n=0}^{\infty} |a_n| \] also converges.
  • This form of convergence guarantees that rearranging the terms of the series does not affect its limit.
  • In simpler terms, if a series converges absolutely, it will also converge normally, but not necessarily vice-versa.
  • Absolute convergence is particularly important in the context of power series because once a series converges absolutely for a value of \( x \), it provides assurance of stability and flexibility in mathematical manipulations, such as term rearrangement.

The condition \( |x| < 1/L \) obtained from the Ratio Test ensures absolute convergence within the radius defined. This highlights the series' behavior in its convergent state, unaffected by individual term changes within the bound.

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

Find the Taylor series of \(f\) about \(a\), and write out the first four terms of the series. $$ f(x)=\left(1+x^{2}\right)^{1 / 3} ; a=0 $$

Express the given series \(\sum_{n=1}^{\infty} a_{n}\) in the form \(c+\sum_{n=4}^{\infty} a_{n}\) $$ \sum_{n=1}^{\infty} \frac{1}{n^{2}} $$

Determine whether the series converges or diverges. $$ \sum_{n=1}^{\infty}(-1)^{n} \cot \left(\frac{\pi}{2}-\frac{1}{n}\right) $$

Determine whether or not the series converges, and if so, find its sum. $$ \sum_{n=0}^{\infty}(-3)\left(\frac{2}{3}\right)^{2 n} $$

Express the repeating decimal as a fraction. $$ 0.72727272 \ldots $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

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.