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Chapter 11: Problem 6

Converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series鈥� convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{-2}{n \sqrt{n}} $$

Short Answer

Expert verified
The series converges by the p-series test.

Step by step solution

01

Analyze the General Term

The general term of the series is \(-\frac{2}{n \sqrt{n}}\). This can be rewritten as \(-\frac{2}{n^{3/2}}\).
02

Check for an Appropriate Test

We consider the p-series test, noting that a p-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges if \( p > 1 \) and diverges if \( p \leq 1 \).
03

Apply the p-Series Test

For the series \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \), the value of \( p \) is \( 3/2 \), which is greater than 1. Hence, this p-series converges.
04

Consider the Sign of the Series

The series involves \(-\frac{2}{n^{3/2}}\), which is negative. Nonetheless, convergence or divergence is determined by the magnitude of terms, which are given by \( \frac{2}{n^{3/2}} \).
05

Conclude the Convergence based on Magnitude

Since the series \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \) converges by the p-series test, the original series \( \sum_{n=1}^{\infty} -\frac{2}{n^{3/2}} \) also converges.

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Key Concepts

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

p-Series Test
A p-series is a specific type of infinite series expressed in the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). The behavior of a p-series largely depends on the value of \( p \). Here are the key points to remember about the p-series test:
  • If \( p > 1 \), the series converges.
  • If \( p \leq 1 \), the series diverges.
In the original problem, the series is \( \sum_{n=1}^{\infty} \frac{-2}{n \sqrt{n}} \), which is reformulated into \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \) without the negative sign for convergence analysis. Here, \( p = 3/2 \), clearly greater than 1, indicating convergence by the p-series test.
Convergence and Divergence
Convergence and divergence are fundamental concepts in the study of infinite series. They help us understand whether a series sums to a finite number or not.
  • A series converges when the sum approaches a finite limit as more terms are added.
  • A series diverges when the sum approaches infinity or does not settle on any value.
For the given series \( \sum_{n=1}^{\infty} \frac{-2}{n \sqrt{n}} \), even though the terms are negative, the analysis focuses on the magnitude of the terms for determining convergence. Since the positive series \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \) converges, the original series is considered convergent in terms of its behavior based on magnitude.
General Term Analysis
The general term of a series provides insight into the series' behavior as \( n \) approaches infinity. For this exercise, the general term is \(-\frac{2}{n \sqrt{n}}\), simplified to \(-\frac{2}{n^{3/2}}\). Breaking it down:
  • The numerator \(-2\) shows that each term in the series is negative, which affects the sign but not convergence.
  • The denominator \(n^{3/2}\) determines the rate at which the series' terms decline. Here, the power \( p = 3/2 \) is crucial for applying the p-series test.
General term analysis, as used in this problem, helps in deciding which test to apply and in predicting the series' convergence behavior based on the form of its terms.
Infinite Series
An infinite series is the sum of infinitely many terms, expressed in the form \( \sum_{n=1}^{\infty} a_n \). Understanding infinite series involves determining whether their sum results in a finite number (converges) or becomes indefinitely large (diverges). A few types of infinite series include geometric series, harmonic series, and the p-series discussed here. In the context of the original exercise, we are dealing with a p-series:
  • The infinite nature of the series implies no finite endpoint, calling for methods like the p-series test to determine behavior.
  • Infinite series necessitate an approach focused on identifying convergence or divergence to understand their sum.
Each type of infinite series has unique properties that dictate which tests or concepts will confirm their convergence or divergence.

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Most popular questions from this chapter

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=\frac{\ln n}{n} $$

Show that the Taylor series for \(f(x)=\tan ^{-1} x\) diverges for \(|x|>1\)

In Exercises \(33-36\) , use series to estimate the integrals' values with an error of magnitude less than \(10^{-3} .\) (The answer section gives the integrals' values rounded to five decimal places.) $$ \int_{0}^{0.1} \frac{1}{\sqrt{1+x^{4}}} d x $$

Use series to evaluate the limits. $$ \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{x} $$

a. Show that $$ \int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \quad(p \text { a positive constant }) $$ converges if and only if \(p>1\) b. What implications does the fact in part (a) have for the convergence of the series $$ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} ? $$ Give reasons for your answer.
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