/*! 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 188 Show that \(\sum_{n=1}^{\infty} ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 188

Show that \(\sum_{n=1}^{\infty} n x^{n-1}\) converges for \(|x|<1\).

Short Answer

Expert verified
The series \( \sum_{n=1}^{\infty} n x^{n-1} \) converges for \( |x|<1 \) as proven by the Root Test.

Step by step solution

01

Rewrite the series as a power series

Notice that the series \( \sum_{n=1}^{\infty} n x^{n-1} \) can be rewritten as \( \sum_{n=0}^{\infty} (n+1) x^n \). This is because when \( n=0 \), the term (n+1)x^n becomes 0, so it doesn't change the sum of the series.
02

Apply the Root Test

The Root Test can be applied to any series in the form \( \sum_{n=0}^{\infty} a_n x^n \). If \( |x| \) times the nth root of \( |a_n| \) tends to a limit as \( n \) approaches infinity, and this limit is less than 1, then the series converges absolutely. So first we calculate the n-th root of \( |a_n| \), which in this case is the n-th root of \( |n+1| \), and this tends to 1 as \( n \) approaches infinity. Since \( |x| \) satisfies \( |x|<1 \), this limit is indeed less than 1, so the series converges.
03

Conclusion

As a result, we have shown using the Root Test that the series \( \sum_{n=1}^{\infty} n x^{n-1} \) converges for \( |x|<1 \).

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.

Root Test
The Root Test is a powerful method used to determine whether an infinite series converges. It's particularly handy when dealing with series of the form \( \sum_{n=0}^{\infty} a_n x^n \). Why is it called the "Root Test"? Because it involves taking the nth root of each term.
Here's how it works: Calculate the limit of the nth root of the absolute value of the terms \( a_n \). Essentially, you're looking for:
  • \( \lim_{{n\to\infty}} \sqrt[n]{|a_n|} \)
If the limit you find is \( L \), then:
  • If \( L < 1 \), the series absolutely converges.
  • If \( L > 1 \) or the limit doesn't exist, the series diverges.
  • If \( L = 1 \), the test is inconclusive.
In the exercise, we calculated the nth root of \( |n+1| \), which approaches 1 as \( n \to \infty \). Since \( |x| < 1 \), the product \( |x| \cdot \lim_{{n\to\infty}} \sqrt[n]{n+1} < 1 \), showing convergence of the series.
Infinite Series
Let's dive into the concept of infinite series, which is at the heart of this problem. An infinite series is a sum of infinitely many numbers arranged in a specific order. Mathematically, it's often presented as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the terms of the series.
The broader classes of infinite series we deal with are:
  • Convergent Series: If the series has a finite sum as you add more terms, it's convergent.
  • Divergent Series: If the sum grows indefinitely or oscillates as you add more terms, it's divergent.
For the exercise at hand, we're interested in determining if our series \( \sum_{n=1}^{\infty} n x^{n-1} \) is convergent. By transforming it, we confirmed that under the condition \( |x| < 1 \), the series converges.
  • Transformation: Notice the shift in terms which simplified applying the Root Test.
This concept is fundamental in calculus and helps understand the behavior of functions defined by power series.
Absolute Convergence
The concept of absolute convergence is crucial when dealing with series. An infinite series \( \sum a_n \) is said to converge absolutely if the series of absolute values \( \sum |a_n| \) converges. Absolute convergence is stronger than regular convergence.
Why is it important? Because:
  • If a series converges absolutely, it converges in the usual sense, and you can rearrange its terms without affecting the sum.
  • This stability makes absolute convergence a robust concept to work with.
In the exercise, applying the Root Test to the series \( \sum_{n=0}^{\infty} (n+1)x^n \) confirmed absolute convergence for \( |x| < 1 \). This means:
  • The magnitude of every term in the series, when summed indefinitely, results in a finite limit.
  • It's resistant to any alternative arrangements of its terms.
Understanding absolute convergence provides assurance about integrating and manipulating series safely and reliably.

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

Observe that for all \(x \in[-1,1]|x| \leq 1 .\) Identify which of the following series converges pointwise and which converges uniformly on the interval \([-1,1] .\) In every case identify the limit function. (a) \(\sum_{n=1}^{\infty}\left(x^{n}-x^{n-1}\right)\) (b) \(\sum_{n=1}^{\infty} \frac{\left(x^{n}-x^{n-1}\right)}{n}\) (c) \(\sum_{n=1}^{\infty} \frac{\left(x^{n}-x^{n-1}\right)}{n^{2}}\) Using the Weierstrass- \(M\) test, we can prove the following result.

Prove the Cauchy criterion. At this point several of the tests for convergence that you probably learned in calculus are easily proved. For example:

Let \(0

Consider the sequence of functions \(\left(f_{n}\right)\) given by $$f_{n}(x)=\left\\{\begin{array}{ll} n & \text { if } x \in\left(0, \frac{1}{n}\right) \\ 0 & \text { otherwise } \end{array}\right.$$ (a) Show that \(f_{n} \stackrel{\text { ptwise }}{\longrightarrow} 0\) on [0,1] , but \(\lim _{n \rightarrow \infty} \int_{x=0}^{1} f_{n}(x) \mathrm{d} x \neq \int_{x=0}^{1} 0 \mathrm{~d} x\). (b) Can the convergence be uniform? Explain.

Prove Theorem 34. $$\int_{t=a}^{x} f_{n}^{\prime}(t) \mathrm{d} t=f_{n}(x)-f_{n}(a)$$ Take the limit of both sides and differentiate with respect to \(x .]\) As before, applying this to power series gives the following result.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

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.