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Chapter 2: Problem 7

Let \(\left\\{x_{n}\right\\}\) be a decreasing sequence such that \(\sum x_{n}\) converges. Show that \(\lim _{n \rightarrow \infty} n x_{n}=0\)

Short Answer

Expert verified
The sequence \( n x_n \) approaches zero because \( x_n \) diminishes faster than \( \frac{1}{n} \).

Step by step solution

01

Understand the sequence behavior

The sequence \( \{x_n\} \) is given to be decreasing, meaning \( x_{n+1} \leq x_n \) for all \( n \). Moreover, the series \( \sum x_n \) is convergent, which means that the terms \( x_n \) must approach zero as \( n \to \infty \).
02

Analyze convergence implications

Since \( \sum x_n \) converges, the terms \( x_n \) approach zero. For a convergent series, the terms must go to zero, i.e., \( \lim_{n\to\infty} x_n = 0 \), though this is not sufficient alone to conclude \( n x_n \rightarrow 0 \).
03

Employ the Cauchy criterion

Given the sequence \( \{x_n\} \) is decreasing and \( \sum x_n \) converges, for any \( \epsilon > 0 \), there exists an \( N \) such that for all \( n > N \), \( x_n < \frac{\epsilon}{n} \). This is a result of the fact that if the terms remained large, the series could not converge.
04

Showing \( n x_n \to 0 \)

For any \( \epsilon > 0 \), select a corresponding \( N \) from the previous step. For all \( n > N \), we have \( x_n < \frac{\epsilon}{n} \). Thus, \( n x_n < \epsilon \). As \( \epsilon \) is arbitrary, this implies that \( \lim_{n\to\infty} n x_n = 0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Decreasing Sequence
In mathematics, a decreasing sequence is a sequence of numbers where each term is less than or equal to the previous term. Decreasing sequences are common in various areas of mathematics and real analysis.
A sequence \( \{x_n\} \) is said to be decreasing if it satisfies the condition \( x_{n+1} \leq x_n \) for all values of \( n \). This property implies that as we progress through the sequence, the terms become smaller or remain the same.
  • Strictly Decreasing: If each subsequent term is strictly less than the previous one, it is known as strictly decreasing sequence (\( x_{n+1} < x_n \)).
  • Application: Decreasing sequences are used to demonstrate various properties in limits and convergence, often simplifying analysis in calculus and real analysis problems.
      • Recognizing the nature of a sequence can help in solving complex problems, especially when combined with convergence principles.
Convergent Series
A convergent series in real analysis is a series whose terms approach a specific value, or limit, as you sum more and more terms. More formally, the infinite series \( \sum x_n \) is considered convergent if the sequence of its partial sums \( S_n = x_1 + x_2 + ... + x_n \) approaches a finite limit.
This implies that as you keep adding more terms to the series, the total does not grow infinitely large or vary wildly. Instead, it settles close to a particular number.
  • Key Feature: In a convergent series, \( \lim_{n\to\infty} S_n = L \, \) where \( L \) is a finite number.
  • Convergence Test: Several tests, such as the ratio test and root test, help determine if a series converges or not. However, for our purposes, knowing that the terms of the series must tend to zero (\( \lim_{n \to \infty} x_n = 0 \)) is crucial.
Convergent series play a pivotal role in many areas of calculus and analysis, providing the foundation for integral calculus and many other mathematical concepts.
Cauchy Criterion
The Cauchy Criterion is a vital tool in understanding series and sequences within real analysis. It provides a way to determine if a sequence converges based on the proximity of its elements as the sequence progresses.
A sequence \( \{a_n\} \) is convergent if for every \( \epsilon > 0 \, \) there exists a positive integer \( N \) such that for all \( m, n > N \, \) the difference \( |a_m - a_n| < \epsilon \.\)
This means that beyond a certain point in the sequence, the terms become arbitrarily close to each other.
  • Application: The Cauchy Criterion is typically used where direct computation of a sequence's limit is challenging.
  • Relation to Series: For series, if the series \( \sum a_n \) is convergent, then the partial sums form a Cauchy sequence.
In the exercise provided, the Cauchy Criterion helps to demonstrate that even though the sequence terms get smaller, they must do so at a pace fast enough to satisfy the condition for convergence of the series.

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Most popular questions from this chapter

Suppose \(\left\\{c_{n}\right\\}\) is any sequence. Prove that for any \(r \in(0,1)\) there exists a strictly increasing sequence \(\left\\{n_{k}\right\\}\) of natural numbers \(\left(n_{k+1}>n_{k}\right)\) such that $$ \sum_{k=1}^{\infty} c_{k} x^{n_{k}} $$ converges absolutely for all \(x \in[-r, r] .\)

Let \(\left\\{x_{n}\right\\}\) be a sequence and define a sequence \(\left\\{y_{n}\right\\}\) by \(y_{2 k}:=x_{k^{2}}\) and \(y_{2 k-1}:=x_{k}\) for all \(k \in \mathbb{N}\). Prove that \(\left\\{x_{n}\right\\}\) converges if and only if \(\left\\{y_{n}\right\\}\) converges. Furthermore, prove that if they converge, then \(\lim x_{n}=\lim y_{n}\).

(Challenging): Let \(\left\\{x_{n}\right\\}\) be a convergent sequence such that \(x_{n} \geq 0\) and \(k \in \mathbb{N}\). Then $$ \lim _{n \rightarrow \infty} x_{n}^{1 / k}=\left(\lim _{n \rightarrow \infty} x_{n}\right)^{1 / k} $$ Hint: Find an expression \(q\) such that \(\frac{x_{n}^{1 / k}-x^{1 / k}}{x_{n}-x}=\frac{1}{q} .\)

a) Suppose \(\left\\{a_{n}\right\\}\) is a bounded sequence and \(\left\\{b_{n}\right\\}\) is a sequence converging to \(0 .\) Show that \(\left\\{a_{n} b_{n}\right\\}\) converges to 0 b) Find an example where \(\left\\{a_{n}\right\\}\) is unbounded, \(\left\\{b_{n}\right\\}\) converges to \(0,\) and \(\left\\{a_{n} b_{n}\right\\}\) is not convergent. c) Find an example where \(\left\\{a_{n}\right\\}\) is bounded, \(\left\\{b_{n}\right\\}\) converges to some \(x \neq 0,\) and \(\left\\{a_{n} b_{n}\right\\}\) is not convergent.

(Easy): Prove the following stronger version of Lemma 2.2.12, the ratio test. Suppose \(\left\\{x_{n}\right\\}\) is a sequence such that \(x_{n} \neq 0\) for all \(n\). a) Prove that if there exists an \(r<1\) and \(M \in \mathbb{N}\) such that for all \(n \geq M\) we have $$ \frac{\left|x_{n+1}\right|}{\left|x_{n}\right|} \leq r $$ then \(\left\\{x_{n}\right\\}\) converges to \(0 .\) b) Prove that if there exists an \(r>1\) and \(M \in \mathbb{N}\) such that for all \(n \geq M\) we have $$ \frac{\left|x_{n+1}\right|}{\left|x_{n}\right|} \geq r $$ then \(\left\\{x_{n}\right\\}\) is unbounded.
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