/*! 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 8 Show that for each integer \(\ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 8

Show that for each integer \(\mathrm{m}>1\), \(\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{m}<\ln m<1+\frac{1}{2}+\ldots+\frac{1}{m-1}\)

Short Answer

Expert verified
Yes, the inequality holds true for all integers m > 1, as proven using mathematical induction and the Taylor series for the natural logarithm function.

Step by step solution

01

Introduction

The given exercise is to prove the following inequality: $$\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{m}<\ln m<1+\frac{1}{2}+\ldots+\frac{1}{m-1}$$ for all integers \(m > 1\). We will use induction to prove this inequality.
02

Base case

Let's check the base case when \(m = 2\): $$\frac{1}{2}<\ln 2<1$$ Since \(\frac{1}{2} = 0.5\) and \(\ln2 \approx 0.6931\), the inequality holds for \(m = 2\). Thus, the base case is valid.
03

Inductive step

Now, let's assume that the inequality holds for some arbitrary integer \(k > 1\): $$\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k}<\ln k<1+\frac{1}{2}+\ldots+\frac{1}{k-1}$$ We want to show that it also holds for \(k + 1\): $$\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k}+\frac{1}{k+1}<\ln (k+1)<1+\frac{1}{2}+\ldots+\frac{1}{k}$$
04

Using the inductive hypothesis

Let's consider the inequality from the inductive step: $$\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k}+\frac{1}{k+1}<\ln (k+1)$$ According to the inductive hypothesis, we can replace the left-side of the inequality with \(\ln k\): $$\ln k+\frac{1}{k+1}<\ln (k+1)$$ To show this, we can use the property of logarithm: \(ln(k + 1) - ln(k) = ln(\frac{k+1}{k})\). Thus, we have to show: $$\frac{1}{k+1}<\ln \left(\frac{k+1}{k}\right)$$
05

Using the Taylor series for the natural logarithm

We can use the Taylor series expansion of \(\ln(1 + x)\) around \(x = 0\) to prove our inequality: $$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$$ Since \(x=\frac{1}{k}\) and \(k>1\), we have: $$\ln \left(\frac{k+1}{k}\right) = \frac{1}{k} - \frac{1}{2k^2} + \frac{1}{3k^3} - \cdots$$ Note that each term in the series is positive and smaller than \(\frac{1}{k+1}\). Therefore, we get: $$\frac{1}{k+1}<\ln \left(\frac{k+1}{k}\right)$$
06

Checking the other side of the inequality

According to the inductive hypothesis, inequality holds for \(k\): $$1 + \frac{1}{2} +\ldots+\frac{1}{k-1}<\ln k$$ Now, we want to show that this inequality holds for \(k+1\): $$1+\frac{1}{2}+\ldots+\frac{1}{k}<\ln (k+1)$$ Since we already proved that \(\frac{1}{k+1}<\ln(\frac{k+1}{k})\), we can write: $$1+\frac{1}{2}+\ldots+\frac{1}{k}< \ln k + \frac{1}{k+1} \leq \ln k + \ln \left(\frac{k+1}{k}\right) = \ln (k+1)$$
07

Conclusion

Now, we have proved that the inequality holds for the integer \(k+1\): $$\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k+1}<\ln (k+1)<1+\frac{1}{2}+\ldots+\frac{1}{k}$$ Thus, by mathematical induction, we have shown that the inequality holds for all integers \(m > 1\).

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!

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

\(\int_{-\infty}^{\infty} \mathrm{f}(\mathrm{x}) \mathrm{dx}\) may not equal \(\lim _{\mathrm{b} \rightarrow \infty} \int_{-\mathrm{b}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} x\) Show that \(\int_{0}^{\infty} \frac{2 \mathrm{xdx}}{\mathrm{x}^{2}+1}\) diverges and hence that \(\int_{-\infty}^{\infty} \frac{2 x d x}{x^{2}+1}\) diverges. Then show that \(\lim _{b \rightarrow \infty} \int_{-b}^{b} \frac{2 x d x}{x^{2}+1}=0\)

Let p be a polynomial of degree atmost 4 such that \(\mathrm{p}(-1)=\mathrm{p}(1)=0\) and \(\mathrm{p}(0)=1\). If \(\mathrm{p}(\mathrm{x}) \leq 1\) for \(x \in[-1,1]\), find the largest value of \(\int^{1} p(x) d x\)

Evaluate the following integrals: (i) \(\int_{-\infty}^{\infty} \frac{x d x}{x^{4}+1}\) (ii) \(\int_{0}^{1} \frac{\ln (1-x)}{x} \mathrm{dx}\) (iii) \(\int_{0}^{\infty} \frac{\mathrm{dx}}{(\mathrm{x}+1)(\mathrm{x}+2)}\) (iv) \(\int_{0}^{\infty} \frac{x^{2} d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}, a, b>0 .\)

Find the sum of the series \(\frac{x^{2}}{1.2}-\frac{x^{3}}{2.3}+\frac{x^{4}}{3.4}-\ldots+(-1)^{n+1} \frac{x^{n+1}}{n(n+1)}+\ldots,|x|<1\)

Which of following integrals are improper ? Why? (a) \(\int_{1}^{2} \frac{1}{2 x-1} \mathrm{dx}\) (b) \(\int_{0}^{1} \frac{1}{2 x-1} d x\) (c) \(\int_{-\infty}^{\infty} \frac{\sin x}{1+x^{2}} d x\) (d) \(\int_{1}^{2} \ln (x-1) d x\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

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.