/*! 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 57 Find the limit of the following ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 57

Find the limit of the following sequences or state that they diverge. $$\left\\{\frac{2 \tan ^{-1} n}{n^{3}+4}\right\\}$$

Short Answer

Expert verified
Answer: The limit of the sequence as n approaches infinity is 0.

Step by step solution

01

Identify the function of the sequence and its limit

The given sequence is defined as follows: $$a_n = \frac{2 \tan^{-1} n}{n^3 + 4}$$ We are to find the limit of this sequence as n approaches infinity: $$\lim_{n \to \infty} a_n$$
02

Use properties of tangent function

Recall that \(\tan^{-1} n\) is the arc-tangent of n and increases monotonically with n as it lies in the interval \((-\frac{\pi}{2},\frac{\pi}{2})\). Therefore, we have: $$\lim_{n \to \infty} \tan^{-1} n = \frac{\pi}{2}$$
03

Compare the sequence with a simpler one

Since \(\tan^{-1} n\) converges to \(\frac{\pi}{2}\), we can compare the given sequence with a simpler one, such as: $$b_n = \frac{2 \cdot \frac{\pi}{2}}{n^3 + 4}$$ Which simplifies to: $$b_n = \frac{\pi}{n^3 + 4}$$ Now, we will find the limit of this new sequence.
04

Find the limit of the simpler sequence

Find the limit of \(b_n\) as n approaches infinity: $$\lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{\pi}{n^3 + 4}$$ As \(n\) approaches infinity, \(n^3 + 4\) increases without bound and the overall fraction goes to zero. Thus, the limit of this simpler sequence is: $$\lim_{n \to \infty} b_n = 0$$
05

Establish and apply the Sandwich Theorem

Notice that for all \(n\), $$0 \leq a_n \leq b_n$$ Since the limit of both \(a_n\) and \(b_n\) as n approaches infinity is 0, we can apply the Sandwich Theorem to conclude that the limit of the given sequence is also 0: $$\lim_{n \to \infty} a_n = 0$$

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

Consider the sequence \(\left\\{x_{n}\right\\}\) defined for \(n=1,2,3, \ldots\) by $$x_{n}=\sum_{k=n+1}^{2 n} \frac{1}{k}=\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2 n}$$ a. Write out the terms \(x_{1}, x_{2}, x_{3}\) b. Show that \(\frac{1}{2} \leq x_{n}<1,\) for \(n=1,2,3, \ldots\) c. Show that \(x_{n}\) is the right Riemann sum for \(\int_{1}^{2} \frac{d x}{x}\) using \(n\) subintervals. d. Conclude that \(\lim _{n \rightarrow \infty} x_{n}=\ln 2\)

Evaluate the limit of the following sequences. $$a_{n}=\frac{n^{8}+n^{7}}{n^{7}+n^{8} \ln n}$$

Evaluate the limit of the following sequences. $$a_{n}=\frac{75^{n-1}}{99^{n}}+\frac{5^{n} \sin n}{8^{n}}$$

Find the limit of the sequence $$\left\\{a_{n}\right\\}_{n=2}^{\infty}=\left\\{\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \cdots\left(1-\frac{1}{n}\right)\right\\}.$$

Two sine series Determine whether the following series converge. a. \(\sum_{k=1}^{\infty} \sin \frac{1}{k}\) b. \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

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.