/*! 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 11 Use the Limit Comparison Test to... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 11

Use the Limit Comparison Test to determine if each series converges or diverges. $$\sum_{n=2}^{\infty} \frac{n(n+1)}{\left(n^{2}+1\right)(n-1)}$$

Short Answer

Expert verified
The series \( \sum_{n=2}^{\infty} \frac{n(n+1)}{(n^2+1)(n-1)} \) diverges.

Step by step solution

01

Identify the given series

The series provided is \( \sum_{n=2}^{\infty} \frac{n(n+1)}{(n^2+1)(n-1)} \). We need to determine if this series converges or diverges.
02

Choose a comparison series

For the Limit Comparison Test, we need to choose a simpler series that resembles the given series for large \( n \). Notice that the expression \( \frac{n(n+1)}{(n^2+1)(n-1)} \) behaves similarly to \( \frac{n^2}{n^3} = \frac{1}{n} \) for large \( n \). Hence, choose \( b_n = \frac{1}{n} \), which is the harmonic series and diverges.
03

Set up the limit comparison

Apply the Limit Comparison Test by computing the limit of the ratio of the given series terms and the comparison series terms as \( n \to fty \): \[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{n(n+1)}{(n^2+1)(n-1)}}{\frac{1}{n}} \].
04

Simplify the limit expression

The limit simplifies to: \[ \lim_{n \to \infty} n \cdot \frac{n(n+1)}{(n^2+1)(n-1)} = \lim_{n \to \infty} \frac{n^3 + n^2}{n^3 - n^2 + n - 1} \].
05

Evaluate the limit

Divide both the numerator and the denominator by \( n^3 \): \[ \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{1 - \frac{1}{n} + \frac{1}{n^2} - \frac{1}{n^3}} \]. As \( n \to \infty \), the limit becomes \( \frac{1 + 0}{1 - 0 + 0 - 0} = 1 \).
06

Conclude using the Limit Comparison Test

The limit calculated is 1, which is positive and finite. According to the Limit Comparison Test, since \( \lim_{n \to \infty} \frac{a_n}{b_n} = 1 \) and \( \sum_{n=2}^{\infty} b_n = \sum_{n=2}^{\infty} \frac{1}{n} \) diverges, the given series \( \sum_{n=2}^{\infty} \frac{n(n+1)}{(n^2+1)(n-1)} \) also diverges.

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!

Key Concepts

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

Series Convergence
Understanding whether a series converges or diverges is crucial in calculus. Series convergence simply means that as you sum more and more terms, the total approaches a definite number. It's like having an endless stack of small papers and as you add more, the height doesn't keep increasing without bound. Instead, it reaches a certain height and stops getting higher. Many tests help us determine convergence, such as the Limit Comparison Test.

The Limit Comparison Test helps us compare a complex series with a simpler one. If they behave similarly, we can conclude they either both converge or both diverge.
  • Choose a simpler series to compare.
  • Compute the limit of the ratio of the series terms.
  • If the limit is positive and finite, both series share the same convergence behavior.
This test is handy because sometimes showing direct convergence is tricky, but comparison simplifies the process.
Divergence
A series diverges if the sum of its terms grows indefinitely. Imagine trying to count an infinite stack of coin flips, where each result adds height. If the stack's height just grows endlessly as you add flips, that's divergence.

Divergence tells us that adding more terms will never stop at a definite amount. Many series, like the harmonic series, naturally diverge.
  • In divergence, the sum of the series becomes unbounded.
  • Even if the terms themselves seem to get smaller, the sum might still grow indefinitely.
  • By using tests like the Limit Comparison Test, one can determine if a series diverges by comparing it to another divergent series.
In essence, divergence indicates a lack of mathematical "balance" in the series sum.
Harmonic Series
The harmonic series is a classic example of a divergent series. It takes the form \( \sum_{n=1}^{\infty} \frac{1}{n} \), meaning you add the reciprocals of all positive numbers.

Here's what's surprising: even though the terms get smaller, the series never converges to a specific number!
  • Terms like \( \frac{1}{2} \), \( \frac{1}{3} \), and so on, keep getting tinier.
  • Yet, the total sum keeps increasing slowly but surely.
  • This divergence is key in comparison tests, making it a "measuring stick" for other series.
The harmonic series often helps us in tests, like the Limit Comparison Test, to establish the divergence of more complex series by comparison.

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

Show that the error \(\left(L-s_{n}\right)\) obtained by replacing a convergent geometric series with one of its partial sums \(s_{n}\) is \(a r^{n} /(1-r)\)

Make up a geometric series \(\sum a r^{n-1}\) that converges to the number 5 if a. \(a=2\) b. \(a=13 / 2\)

Pythagorean triples \(\quad\) A triple of positive integers \(a, b,\) and \(c\) is called a Pythagorean triple if \(a^{2}+b^{2}=c^{2} .\) Let \(a\) be an odd positive integer and let $$b=\left\lfloor\frac{a^{2}}{2}\right\rfloor \quad \text { and } \quad c=\left\lceil\frac{a^{2}}{2}\right\rceil$$ be, respectively, the integer floor and ceiling for \(a^{2} / 2\). a. Show that \(a^{2}+b^{2}=c^{2} .\) (Hint: Let \(a=2 n+1\) and express \(b \text { and } c \text { in terms of } n .)\) b. By direct calculation, or by appealing to the accompanying figure, find $$\lim _{a \rightarrow \infty} \frac{\left\lfloor\frac{a^{2}}{2}\right\rfloor}{\left\lceil\frac{a^{2}}{2}\right\rceil}.$$

The first term of a sequence is \(x_{1}=1 .\) Each succeeding term is the sum of all those that come before it: $$x_{n+1}=x_{1}+x_{2}+\cdots+x_{n}$$ Write out enough early terms of the sequence to deduce a general formula for \(x_{n}\) that holds for \(n \geq 2\).

It is not yet known whether the series $$ \sum_{n=1}^{\infty} \frac{1}{n^{3} \sin ^{2} n} $$converges or diverges. Use a CAS to explore the behavior of the series by performing the following steps. a. Define the sequence of partial sums $$ s_{k}=\sum_{n=1}^{k} \frac{1}{n^{3} \sin ^{2} n} $$ What happens when you try to find the limit of \(s_{k}\) as \(k \rightarrow \infty ?\) Does your CAS find a closed form answer for this limit? b. Plot the first 100 points \(\left(k, s_{k}\right)\) for the sequence of partial sums. Do they appear to converge? What would you estimate the limit to be? c. Next plot the first 200 points \(\left(k, s_{k}\right) .\) Discuss the behavior in your own words. d. Plot the first 400 points \(\left(k, s_{k}\right) .\) What happens when \(k=355 ?\) Calculate the number \(355 / 113 .\) Explain from you calculation what happened at \(k=355 .\) For what values of \(k\) would you guess this behavior might occur again?
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Logic and Functions

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.