/*! 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 1 The sequence \((1 / \sqrt{n})\) ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 1

The sequence \((1 / \sqrt{n})\) has limit 0. For each of \(\epsilon=0.01,0.001,0.0001\), determine an integer \(N\) with the property that \(|(1 / \sqrt{n})-0|<\epsilon\) for all \(n>N\)

Short Answer

Expert verified
\(N = 10000\) for \(\epsilon = 0.01\), \(N = 1000000\) for \(\epsilon = 0.001\), \(N = 100000000\) for \(\epsilon = 0.0001\).

Step by step solution

01

Understand the Problem

We need to find an integer \(N\) for each given \(\epsilon\) such that \(\left|\frac{1}{\sqrt{n}} - 0\right| < \epsilon\) for all \(n > N\). This means we need to make \(\frac{1}{\sqrt{n}}\) less than a small positive \(\epsilon\).
02

Set Up the Inequality

For \(\left|\frac{1}{\sqrt{n}}\right| < \epsilon\), we simplify it to \(\frac{1}{\sqrt{n}} < \epsilon\). This implies \(\sqrt{n} > \frac{1}{\epsilon}\).
03

Solve for \(n\)

Square both sides of the inequality \(\sqrt{n} > \frac{1}{\epsilon}\), resulting in \(n > \frac{1}{\epsilon^2}\). This gives us an inequality for \(n\).
04

Determine \(N\) Given \(\epsilon = 0.01\)

For \(\epsilon = 0.01\), calculate \(\frac{1}{0.01^2} = 10000\). Thus, \(N = 10000\) because for all \(n > 10000\), \(\frac{1}{\sqrt{n}} < 0.01\).
05

Determine \(N\) Given \(\epsilon = 0.001\)

For \(\epsilon = 0.001\), calculate \(\frac{1}{0.001^2} = 1000000\). Therefore, \(N = 1000000\) as \(\frac{1}{\sqrt{n}} < 0.001\) for all \(n > 1000000\).
06

Determine \(N\) Given \(\epsilon = 0.0001\)

For \(\epsilon = 0.0001\), calculate \(\frac{1}{0.0001^2} = 100000000\). So, \(N = 100000000\) ensuring \(\frac{1}{\sqrt{n}} < 0.0001\) for \(n > 100000000\).

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Key Concepts

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

Limit of a Sequence
The limit of a sequence is a fundamental concept in calculus and mathematical analysis. It helps us understand the behavior of a sequence as it progresses towards infinity. When we say that a sequence has a limit, we mean that as we go further and further in the sequence, the terms get closer and closer to a specific number. This specific number is called the "limit." For instance, the sequence \((1/\sqrt{n})\) has its terms getting smaller and approaching zero as \(n\) becomes very large, hence we say the limit is 0.

Understanding the limit requires:
  • Recognizing the pattern of the sequence.
  • Determining how terms change as \(n\) increases.
  • Checking if the terms get progressively nearer to a fixed point.
For the sequence \((1/\sqrt{n})\), as \(n\) increases, \( \sqrt{n} \) becomes larger, making the fraction \(1/\sqrt{n}\) smaller, consistently pushing the values towards 0.
Epsilon-Delta Definition
The epsilon-delta definition is a rigorous approach to formally define what it means for a sequence to approach a limit. It is a method that uses two parameters, \(\epsilon\) and \(\delta\), to describe how close the terms of a sequence can get to its limit. This definition is integral for proving that a sequence indeed converges to its limit.

In the epsilon-delta framework:
  • \(\epsilon\) represents how close we want the sequence to be to the limit.
  • We find a \(\delta\) (or in sequence terms, an integer \(N\)) so that for all terms beyond this \(N\), the difference between the sequence term and the limit is less than \(\epsilon\).
Applying this to our example \((1/\sqrt{n})\), for a specific \(\epsilon\), we calculate the smallest \(N\) such that for all \(n > N\), \(|1/\sqrt{n} - 0| < \epsilon\). This ensures that the terms are sufficiently close to 0, reflecting the sequence's convergence towards its limit.
Asymptotic Behavior
Asymptotic behavior describes how a sequence behaves as the index \(n\) becomes very large. It's a way to examine the end behavior of a sequence, which often involves looking at how closely a sequence approaches a limit or an asymptotic line as \(n\) approaches infinity.

Key aspects of asymptotic behavior:
  • Determines the rate at which a sequence approaches its limit.
  • Essential for comparing sequences or functions based on their growth or decay.
For example, in our given sequence \((1/\sqrt{n})\), as \(n\) tends to infinity, the sequence approaches 0. We describe its asymptotic behavior by saying it behaves like 0 as \(n\) grows, because its terms diminish and get arbitrarily close to zero. Understanding this behavior is crucial when we need to assess how quickly the terms approach their limit, which can be informative in advanced mathematical and real-world applications.

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Most popular questions from this chapter

Let \(\sum_{n=1}^{\infty} a_{n}\) and \(\sum_{n=1}^{\infty} b_{n}\) be convergent series of non-negative terms. Show that \(\sum_{n=1}^{\infty}\left(a_{n} b_{n}\right)^{1 / 2}\) is convergent. Give an example to show that the converse implication is false.

A celebrated theorem due to Riemann \(^{8}\) shows that a conditionally convergent series can be rearranged so as to sum to any real number, or to diverge to \(\infty\), or to diverge to \(-\infty\). This exercise has the more modest aim of showing that a rearrangement may have a different sum. Consider the alternating harmonic series $$ 1-\frac{1}{2}+\frac{1}{3}-\cdots $$ with sum \(S\), and denote its sum to \(n\) terms \((n=1,2,3, \ldots)\) by \(S_{n}\). Consider also the rearranged series $$ 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\cdots $$ and denote its sum to \(n\) terms by \(T_{n}\). For each \(n \geq 1\), let $$ H_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} $$ a) Show that \(S_{2 n}=H_{2 n}-H_{n}\) for all \(n \geq 1\). b) Show that $$ T_{3 n}=H_{4 \mathrm{n}}-\frac{1}{2} H_{2 n}-H_{n}=S_{4 n}+\frac{1}{2} S_{2 n} $$ and deduce that the rearranged series has sum \(3 S / 2\).

Determine the limit of \(\left(a_{n}\right)\), where $$ a_{1}=2, \quad a_{n+1}=\sqrt{2+2 a_{n}} \quad(n \geq 1) $$

Consider the sequence \(\left(a^{1 / n}\right)\), where \(a>1\). Use the binomial theorem. to show that \(a>1+n h_{n}\), where \(h_{n}=a^{1 / n}-1\). Deduce that \(\left(a^{1 / n}\right) \rightarrow\) 1. Show that this holds also for \(0

Let \(\left(a_{n}\right),\left(x_{n}\right)\) and \(\left(b_{n}\right)\) be sequences with limits \(\alpha, \xi, \beta\), respectively, and suppose that, for all \(n \geq 1\), $$ a_{n} \leq x_{n} \leq b_{n} $$ Show that \(\alpha \leq \xi \leq \beta\). [This is sometimes called the sandwich principle. It can be useful if \(\left(a_{n}\right)\) and \(\left(b_{n}\right)\) are "known" sequences and \(\left(x_{n}\right)\) is unknown. It is especially useful when \(\alpha=\beta\), for in this case we conclude that \(\xi=\alpha=\beta\).]
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