/*! 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 12 Determine convergence or diverge... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine convergence or divergence of the series. $$\sum_{k=0}^{\infty} \frac{\sqrt{k}}{k^{2}+1}$$

Short Answer

Expert verified
The series \(\sum_{k=0}^{\infty} \frac{\sqrt{k}}{k^{2}+1}\) diverges.

Step by step solution

01

Specify the series to compare

The series to compare with is \(\frac{1}{k}\). We are familiar that the harmonic series \(\sum_{k=1}^{\infty} \frac{1}{k}\) diverges.
02

Perform the Comparison Test

To apply the Comparison Test, we need to compare the terms of our given series with the terms from the series we selected in Step 1. As \(\frac{\sqrt{k}}{k^{2}+1} \le \frac{1}{k}\), we deduce that our series is smaller than the divergent series \(\sum_{k=1}^{\infty} \frac{1}{k}\). According to the Comparison Test, if a series is smaller than a divergent series, then it also diverges.
03

Conclude the Result

Based on the Comparison Test, we conclude that the given series \(\sum_{k=0}^{\infty} \frac{\sqrt{k}}{k^{2}+1}\) 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 91Ó°ÊÓ!

Key Concepts

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

Comparison Test
The Comparison Test is an essential tool for determining whether a series converges or diverges. It involves comparing a given series to a second series with known convergence properties.
Simply, if you want to know if the series \( \sum a_k \) converges, you compare it to a series \( \sum b_k \) that you already know. Here's how it works:
  • If \( 0 \le a_k \le b_k \) for all \( k \) and \( \sum b_k \) converges, then \( \sum a_k \) also converges.
  • If \( a_k \ge b_k \ge 0 \) for a sufficient number of terms and \( \sum b_k \) diverges, then \( \sum a_k \) also diverges.
In our example, the series \( \sum_{k=0}^{\infty} \frac{\sqrt{k}}{k^{2}+1} \) is compared with the harmonic series \( \sum_{k=1}^{\infty} \frac{1}{k} \). Since \( \frac{\sqrt{k}}{k^{2}+1} \le \frac{1}{k} \), and the harmonic series is known to diverge, the Comparison Test tells us our original series also diverges. This test is powerful as it allows researchers and students to determine the behavior of complex series by relating them to simpler, well-known series.
Harmonic Series
The harmonic series is one of the first series many students encounter that diverges. It takes the form \( \sum_{k=1}^{\infty} \frac{1}{k} \).
At first glance, it might seem counterintuitive that this series diverges—after all, its terms get progressively smaller the further you go along. However, it is important to note that the sum of these ever-decreasing terms keeps increasing without bound.
In other words, no matter how far you go, the total sum never settles at a finite value.
  • The essence of divergence in the harmonic series is due to its slow decay rate, meaning the terms decrease too slowly.
  • This series serves as a common comparator in the Comparison Test.
This is why it appears in our textbook example, as a benchmark for determining the behavior of other series.
Divergent Series
A divergent series is one that does not have a finite limit. As you add more terms, the total sum of the series either becomes infinitely large or doesn't approach a single finite number.
In mathematical analysis, understanding divergence is crucial, as it tells us the behavior of functions over an infinite domain.
  • Divergent series may appear deceivingly similar to convergent series, as their terms also get smaller. However, the rate and manner of decay differ significantly.
  • Often, divergent series like the harmonic series are utilized to test other series' convergence or divergence properties via techniques such as the Comparison Test.
In the case of \( \sum_{k=0}^{\infty} \frac{\sqrt{k}}{k^{2}+1} \), divergence was shown using the Comparison Test by linking it to the known divergent harmonic series. This tells us that while the terms might trend towards zero, the series sum itself doesn't converge to a finite limit.

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

For any constants \(a\) and \(b>0,\) determine the interval and radius of convergence of \(\sum_{k=0}^{\infty} \frac{(x-a)^{k}}{b^{k}}\)

Define \(a_{n}=\frac{1}{n^{2}}+\frac{2}{n^{2}}+\cdots+\frac{n}{n^{2}} .\) Evaluate the sum using a formula from section 4.2 and show that the sequence converges. By thinking of \(a_{n}\) as a Riemann sum, identify the definite integral to which the sequence converges.

Determine the radius and interval of convergence. $$\sum_{k=2}^{\infty} k^{2}(x-3)^{k}$$

This problem is sometimes called the coupon collectors" problem. The problem is faced by collectors of trading cards. If there are \(n\) different cards that make a complete set and you randomly obtain one at a time, how many cards would you expect to obtain before having a complete set? (By random, we mean that each different card has the same probability of \(\frac{1}{n}\) of being the next card obtained.) In exercises \(69-72,\) we find the answer for \(n=10 .\) The first step is simple; to collect one card you need to obtain one card. Now, given that you have one card, how many cards do you need to obtain to get a second (different) card? If you're lucky, the next card is it (this has probability 10). But your next card might be a duplicate, then you get a new card (this has probability \(\left.\frac{1}{10} \cdot \frac{9}{10}\right) .\) Or you might get two duplicates and then a new card (this has probability \(\left.\frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10}\right) ;\) and so on. The mean is \(1 \cdot \frac{9}{10}+2 \cdot \frac{1}{10} \cdot \frac{9}{10}+3 \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10}+\cdots\) or \(\sum_{k=1}^{\infty} k\left(\frac{1}{10}\right)^{k-1}\left(\frac{9}{10}\right)=\sum_{k=1}^{\infty} \frac{9 k}{10^{k}} .\) Using the same trick as in exercise \(68,\) show that this is a convergent series with \(\operatorname{sum} \frac{10}{9}\)

The function \(f(x)=\left\\{\begin{array}{ll}-1, & \text { if }-2

See all solutions

Recommended explanations on Math Textbooks

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.