/*! 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 7 State whether each statement is ... [FREE SOLUTION] | 91影视

91影视

State whether each statement is true, or give an example to show that it is false. $$ \text { If } \sum_{n=1}^{\infty} a_{n} x^{n} \text { converges, then } a_{n} x^{n} \rightarrow 0 \text { as } n \rightarrow \infty \text { . } $$

Short Answer

Expert verified
The statement is false; convergence of a series doesn't imply each term goes to zero.

Step by step solution

01

Understanding the Problem

We need to determine if the statement is true or provide a counterexample. The statement claims that if the power series \( \sum_{n=1}^{\infty} a_n x^n \) converges, then \( a_n x^n \to 0 \) as \( n \to \infty \).
02

Reviewing Convergence of Series

For a series \( \sum_{n=1}^{\infty} a_n x^n \) to converge, it means that the partial sums \( S_N = a_1 x + a_2 x^2 + \ldots + a_N x^N \) approach a finite limit as \( N \rightarrow \infty \).
03

Examining Terms of a Convergent Series

In general, for the terms of a series \( a_n x^n \), convergence of the series does not necessarily imply that \( a_n x^n \rightarrow 0 \). Convergence is a property of the entire sum, not necessarily of the individual terms approaching zero.
04

Providing a Counterexample

Consider the geometric series \( \sum_{n=1}^{\infty} x^n \) with \( x = \frac{1}{2} \). This series converges to \( 1 \). However, \( a_n x^n = x^n = \left( \frac{1}{2} \right)^n \). As \( n \) increases, \( \left( \frac{1}{2} \right)^n \rightarrow 0 \), yet this is not always the case in general for all convergent series unless the radius of convergence is less than 1 for the entire series to converge absolutely.
05

Conclusion Evaluation

The assumption that \( a_n x^n \rightarrow 0 \) from convergence of the entire series is not universally valid across all series. Hence, the statement is false, generally proven by the example of a geometric series with \( |x| < 1 \).

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.

Power Series
A power series is an infinite series of the form \( \sum_{n=0}^{\infty} a_n x^n \), where \( a_n \) represents the coefficients of the series and \( x \) is a variable. Each term of the series involves a power of \( x \), hence the name 'power series.'
Power series can represent functions and are frequently used in calculus and analysis. They allow us to express complex functions in simpler polynomial forms.
Understanding a power series involves knowing how its terms behave as more terms are added. The ability of a power series to represent a function depends significantly on its convergence properties.
When dealing with power series, one must consider the "radius of convergence," which determines the interval within which the series converges. Outside this interval, the series may diverge.
Geometric Series
Geometric series are a special type of series where each term is a constant multiple (called the common ratio, \( r \)) of the preceding term. A geometric series takes the general form \( a + ar + ar^2 + ar^3 + \ldots \).
An example is the series \( 1 + \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + \ldots \). This series is geometric with a common ratio of \( \frac{1}{2} \).
There is a simple criterion for convergence: a geometric series converges if and only if the absolute value of its common ratio is less than 1, i.e., \( |r| < 1 \). The sum of an infinite geometric series is given by the formula \( \frac{a}{1-r} \) where \( a \) is the first term.
Understanding geometric series helps in recognizing convergence patterns and is beneficial when dealing with more complex series problems.
Partial Sums
When studying series, partial sums help in understanding how the series behaves. A partial sum of a series \( \sum_{n=1}^{\infty} a_n \) is the sum of its first \( N \) terms: \( S_N = a_1 + a_2 + \ldots + a_N \).
Partial sums are used to analyze whether the series converges. If the sequence of partial sums \( S_1, S_2, S_3, \ldots \) approaches a definite limit as \( N \) increases, the series converges.
This concept is crucial because convergence of a series is characterized by the behavior of its partial sums, rather than the individual terms approaching zero. A series might converge even if its terms do not get closer to zero, as illustrated by some geometric series.
Radius of Convergence
In the context of power series, the radius of convergence is a key concept. It defines the range of values for \( x \) for which the series \( \sum_{n=0}^{\infty} a_n x^n \) converges.
To determine the radius of convergence \( R \), one can use the root test or ratio test. The series converges absolutely when \( |x| < R \) and diverges when \( |x| > R \).
The radius of convergence is important as it also indicates the nature of convergence inside, on, and outside the given radius. Within \( |x| < R \), convergence is guaranteed, but on the boundary \( |x| = R \), the behavior must be individually evaluated.
Understanding the radius of convergence provides a framework for using power series to approximate functions within the interval specified by \( R \).

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

In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of f. $$ f(x)=\ln (\cos x) $$

In the following exercises, find the Taylor series of the given function centered at the indicated point. $$ \left.F(x)=\int_{0}^{x} \cos (\sqrt{t}) d t ; f(t)=\sum_{n=0}^{\infty}(-1)^{n} \frac{t^{n}}{(2 n) !} \text { at } a=0 \text { (Note: } f \text { is the Taylor series of } \cos (\sqrt{t}) .\right) $$

Evaluate \(\int_{0}^{\pi / 2} \sin ^{4} \theta d \theta\) in the approximation \(T=4 \sqrt{\frac{L}{g}} \int_{0}^{\pi / 2}\left(1+\frac{1}{2} k^{2} \sin ^{2} \theta+\frac{3}{8} k^{4} \sin ^{4} \theta+\cdots\right) d \theta\) to obtain an improved estimate for \(T\).

Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less than \(\frac{1}{100}\) Compare the accuracy of the polynomial integral estimate with the remainder estimate. $$ \begin{aligned} &\text { [T] } \int_{0}^{2} e^{-x^{2}} d x ; p_{11}=1-x^{2}+\frac{x^{4}}{2}-\frac{x^{6}}{3 !}+\cdots-\frac{x^{22}}{11 !} \text { (You may assume that the absolute value of the } 23 \mathrm{rd}\\\ &\text { derivative of } e^{-x^{2}} \text { is less than } 2 \times 10^{14} \text { .) } \end{aligned} $$

Suppose that \(\sum_{n=0}^{\infty} a_{n} x^{n}\) converges to a function \(f(x)\) such that \(f(0)=1, f^{\prime}(0)=0\), and \(f^{\prime \prime}(x)=-f(x)\). Find a formula for \(a_{n}\) and plot the partial sum \(S_{N}\) for \(N=20\) on \([-5,5]\).

See all solutions

Recommended explanations on Math Textbooks

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.