/*! 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 34 Find the limit of the following ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 34

Find the limit of the following sequences or determine that the limit does not exist. $$\left\\{\frac{(-1)^{n+1} n^{2}}{2 n^{3}+n}\right\\}$$

Short Answer

Expert verified
$$\lim_{n\to\infty} \frac{(-1)^{n+1}n^2}{2n^3 + n}$$ Answer: The limit of the given sequence as n approaches infinity is 0.

Step by step solution

01

Simplify the Expression

Simplify the expression inside the sequence by factoring out common terms from the numerator and the denominator. $$\frac{(-1)^{n+1} n^{2}}{2 n^{3}+n} = \frac{(-1)^{n+1}n^2}{n(2n^2 + 1)}$$
02

Factor out n

Now, we can factor out n from both the numerator and the denominator to get: $$\frac{(-1)^{n+1}n^2}{n(2n^2 + 1)} = \frac{(-1)^{n+1} \times n \times n}{n \times (2n^2 + 1)}$$ Cancelling out the common factor n, we get: $$\frac{(-1)^{n+1}n}{2n^2 + 1}$$
03

Determine the Dominant Term

Next, we determine the dominant terms in the numerator and denominator. As n approaches infinity, (-1)^{n+1} oscillates between -1 and 1, so it does not affect the limit significantly. The dominant term in the numerator is n and the dominant term in the denominator is 2n^2.
04

Calculate the Limit

Now we calculate the limit as n approaches infinity: $$\lim_{n\to\infty} \frac{(-1)^{n+1}n}{2n^2 + 1} = \lim_{n\to\infty} \frac{n}{2n^2 + 1}$$ As n approaches infinity, the term 2n^2 in the denominator grows much faster compared to n in the numerator. Therefore, the limit will approach zero. So, we have: $$\lim_{n\to\infty} \frac{(-1)^{n+1}n}{2n^2 + 1} = 0$$ The limit of the given sequence exists and is equal to 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

Convergence parameter Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}$$

Consider the following situations that generate a sequence. a. Write out the first five terms of the sequence. b. Find an explicit formula for the terms of the sequence. c. Find a recurrence relation that generates the sequence. d. Using a calculator or a graphing utility, estimate the limit of the sequence or state that it does not exist. The Consumer Price Index (the CPI is a measure of the U.S. cost of living) is given a base value of 100 in the year \(1984 .\) Assume the CPI has increased by an average of \(3 \%\) per year since \(1984 .\) Let \(c_{n}\) be the CPI \(n\) years after \(1984,\) where \(c_{0}=100.\)

Given that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6},\) show that \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{2}}=\frac{\pi^{2}}{12}.\) (Assume the result of Exercise 63.)

Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty} x^{2 k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

Reciprocals of odd squares Assume that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Mechanics 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.