/*! 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 19 Determine convergence or diverge... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine convergence or divergence of the series. $$\sum_{k=0}^{\infty} \frac{4}{\sqrt{k^{3}+1}}$$

Short Answer

Expert verified
Given that the reference series converges, and according to the Comparison Test our series also converges.

Step by step solution

01

Identify the series structure

Look at the series \(\sum_{k=0}^{\infty} \frac{4}{\sqrt{k^{3}+1}}\). Note that the series is a positive term series
02

Choose a reference series for comparison

For a comparison, we can choose the series \(\sum_{k=0}^{\infty} \frac{1}{k^{\frac{3}{2}}}\). It is known that this type of series, named \(p\)-series, converges if \(p>1\)
03

Apply the Comparison Test

The Comparison Test states that if \(0\le a_{k}\le b_{k}\) for all \(k\) and \(\sum_{k=0}^{\infty} b_{k}\) converges, then \(\sum_{k=0}^{\infty} a_{k}\) also converges. Here, we let \(a_{k}=\frac{4}{\sqrt{k^{3}+1}}\) and \(b_{k}= \frac{4}{k^{\frac{3}{2}}}\). You can quickly see that \(a_{k} \le b_{k}\) for all \(k\ge 0\). So, if the series for \(b_{k}\) converges, the series for \(a_{k}\) will too.
04

Analyze convergence of reference series

In step 2, we identified the reference series as \(\sum_{k=0}^{\infty} \frac{4}{k^{\frac{3}{2}}}\) which can be seen to be a \(p\)-series with \(p=\frac{3}{2}>1\). Therefore, by the p-Series Convergence Test, this series converges.

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Key Concepts

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

Comparison Test
The Comparison Test is a handy tool for determining the convergence of positive term series. It involves comparing a given series to a reference series whose convergence is known. The basic idea is to find another series, say \( \sum b_k \), that is easier to evaluate, and still provides useful information about the original series, \( \sum a_k \).
  • If \(0 \leq a_k \leq b_k\) for all natural numbers \(k\), and \( \sum b_k \) is known to converge, then \( \sum a_k \) also converges.
  • If \( \sum b_k \) diverges and \( a_k \geq b_k \), then \( \sum a_k \) also diverges.
By establishing an inequality between the original series and the reference series, we can draw conclusions about the convergence behavior without directly calculating the original series.
p-series
A \(p\)-series takes the form \( \sum_{k=1}^{\infty} \frac{1}{k^p} \). This type of series is particularly useful as the basis for comparison tests due to its well-known convergence criteria. Specifically, a \(p\)-series converges if \(p > 1\) and diverges if \(p \leq 1\).
In the solution to our problem, the comparison series \( \sum_{k=0}^{\infty} \frac{1}{k^{3/2}} \) is a \(p\)-series with \(p = \frac{3}{2}\) which is greater than 1, ensuring convergence. This makes it a perfect candidate to apply the Comparison Test, as it confirms the convergence of our main series without tedious computations.
Understanding \(p\)-series fundamentals is crucial as these series often emerge in both mathematical theory and practical applications, serving as a foundation for analyzing the behavior of more complex series.
Convergence and Divergence
Series convergence refers to the sum of the terms tending to a specific finite value as the number of terms approaches infinity. Conversely, divergence means the sum grows indefinitely without approaching a finite limit.For series analysis, determining convergence or divergence allows us to understand the long-term behavior of the series. This involves using tests and methods, like the Comparison Test, and identifying properties of the series such as whether it is a \(p\)-series.
  • Convergent series have well-defined limits and finite sums.
  • Divergent series do not settle to a definite value, implying the sum grows too large as more terms are added.
In our exercise, the identified convergence of the comparison series led directly to the conclusion that the original series also converges, thanks to the Comparison Test and the nature of \(p\)-series.
Positive Term Series
A series with positive terms is one where all terms are positive numbers. This characteristic is crucial in the application of various convergence tests, such as the Comparison Test.
  • For positive term series, ordering terms from smallest to largest is straightforward, which simplifies comparisons.
  • The positivity condition ensures that any partial sum of the series is non-decreasing, providing a clearer path to analyze convergence.
In this exercise's original series \( \sum_{k=0}^{\infty} \frac{4}{\sqrt{k^3+1}} \), all terms are positive, making it a suitable candidate for using the Comparison Test. This simplifies the problem, allowing us to identify an appropriate reference series and directly apply convergence criteria.

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Most popular questions from this chapter

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 \(\operatorname{sum} \frac{10}{9}\)

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