/*! 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 26 Converge, and which diverge? Use... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 26

Converge, and which diverge? Use any method, and give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^{3}+2}}$$

Short Answer

Expert verified
The series converges by the Comparison Test.

Step by step solution

01

Identify the Series Type

The given series is an infinite series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n^{3}+2}} \). This looks like a series that can be tested using the Comparison Test or Limit Comparison Test for convergence.
02

Simplify the General Term

Observe that the term \( \frac{1}{\sqrt{n^{3}+2}} \approx \frac{1}{\sqrt{n^{3}}} = \frac{1}{n^{3/2}} \) for large values of \( n \). The main idea is to compare the given series with \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \).
03

Use the Comparison Test

We know that \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \) converges, because it is a \( p \)-series with \( p = 3/2 > 1 \), which means it converges. Now, to use the comparison test: \[\frac{1}{\sqrt{n^3 + 2}} \leq \frac{1}{n^{3/2}}\] for all \( n \geq 1 \). Since the series \( \sum \frac{1}{n^{3/2}} \) converges, by the Comparison Test, \( \sum \frac{1}{\sqrt{n^{3}+2}} \) also converges.
04

Confirm the Conclusion

Thus, since the comparison to a known convergent \( p \)-series is valid, the infinite series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n^{3}+2}} \) converges by the Comparison Test.

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.

Comparison Test
The Comparison Test is a powerful method to determine whether an infinite series converges or diverges. The idea is simple: compare our series to another series whose convergence properties we already know.

To use the Comparison Test effectively, follow these steps:
  • Choose a comparison series that resembles the series in question. This series should have known convergence or divergence characteristics.
  • Check if the terms of the series of interest are always less than or equal to those of the comparison series.
  • If our series has terms smaller than a convergent comparison series, it too will converge. Conversely, if the terms are greater than a divergent series, then our series diverges.
For example, if we apply this to the series in the problem, we choose the series \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \). This series is known to converge, and hence our original series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n^{3}+2}} \), which has terms smaller than the comparison series, will also converge by the Comparison Test.
Limit Comparison Test
The Limit Comparison Test is another technique for series convergence testing, similar to the regular Comparison Test but often more flexible. This test works well when the Comparison Test is not straightforward to use.

Here鈥檚 how the Limit Comparison Test works:
  • Select a standard series that is comparable to the given series, usually a p-series is handy for this.
  • Calculate the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \), where \( a_n \) represents the general term of our series and \( b_n \) represents the term from the comparison series.
  • If this limit is a positive finite number, both series will either converge or diverge together.
Understanding this test provides a robust tool when confronted with series whose behavior isn't entirely clear, enabling a deeper mathematical insight.
p-series
A p-series is a specific type of infinite series that follows the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). Whether the series converges or diverges can be determined by the value of \( p \). Here鈥檚 the key:
  • If \( p > 1 \), the series converges.
  • If \( p \leq 1 \), the series diverges.
In many problems, p-series act as a useful comparator, as they have well-defined convergence properties. For our exercise, the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \) was used because \( p = 3/2 \) is greater than 1, proving its convergence. Hence, this series provides a useful baseline to compare with more complicated terms.
Infinite Series
When studying infinite series, we are fundamentally dealing with a sum that has an infinite number of terms. The main question we ask is whether this infinite sum has a finite value, which means it converges, or if it grows without bound, indicating divergence.

To analyze these, various tests are applied based on the nature of the series:
  • Comparison Test and Limit Comparison Test, as discussed, are pivotal in assessing convergence.
  • Other tests include the Ratio Test, Root Test, and more, each useful in different scenarios.
Infinite series occur frequently in calculus and real analysis, playing a crucial role in understanding functions and solving complex problems. The ability to discern whether an infinite series converges or diverges is a foundational skill in advanced mathematics.

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

The Taylor polynomial of order 2 generated by a twice-differentiable function \(f(x)\) at \(x=a\) is called the quadratic approximation of \(f\) at \(x=a\). In Exercises \(45-50,\) find the (a) linearization (Taylor polynomial of order 1 ) and (b) quadratic approximation of \(f\) at \(x=0.\) $$f(x)=\sin x$$

Use a geometric series to represent each of the given functions as a power series about \(x=0,\) and find their intervals of convergence. a. \(f(x)=\frac{5}{3-x}\) b. \(g(x)=\frac{3}{x-2}\)

Series for \(\tan ^{-1} x\) for \(|x|>1\) Derive the series $$\begin{array}{l}\tan ^{-1} x=\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x>1 \\\\\tan ^{-1} x=-\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x<-1\end{array}$$ by integrating the series $$\frac{1}{1+t^{2}}=\frac{1}{t^{2}} \cdot \frac{1}{1+\left(1 / t^{2}\right)}=\frac{1}{t^{2}}-\frac{1}{t^{4}}+\frac{1}{t^{6}}-\frac{1}{t^{8}}+\cdots$$ in the first case from \(x\) to \(\infty\) and in the second case from \(-\infty\) to \(x\).

Uniqueness of convergent power series a. Show that if two power series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) and \(\sum_{n=0}^{\infty} b_{n} x^{n}\) are convergent and equal for all values of \(x\) in an open interval \((-c, c),\) then \(a_{n}=b_{n}\) for every \(n .\) (Hint: Let \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}=\sum_{n=0}^{\infty} b_{n} x^{n} .\) Differentiate term by term to show that \(\left.a_{n} \text { and } b_{n} \text { both equal } f^{(n)}(0) /(n !) .\right)\) b. Show that if \(\sum_{n=0}^{\infty} a_{n} x^{n}=0\) for all \(x\) in an open interval \((-c, c),\) then \(a_{n}=0\) for every \(n\).

a. Find the interval of convergence of the power series $$\sum_{n=0}^{\infty} \frac{8}{4^{n+2}} x^{n}$$ b. Represent the power series in part (a) as a power series about \(x=3\) and identify the interval of convergence of the new series. (Later in the chapter you will understand why the new interval of convergence does not necessarily include all of the numbers in the original interval of convergence.)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Decision 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.