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Chapter 9: Problem 14

Classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{5 n^{1.1}} $$

Short Answer

Expert verified
The series is absolutely convergent.

Step by step solution

01

Understand the Type of Series

The given series is \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{5 n^{1.1}} \). This is an alternating series because it has the term \((-1)^{n+1}\), which causes the terms to alternate in sign.
02

Determine Convergence with Alternating Series Test

The Alternating Series Test stipulates that an alternating series \( \sum (-1)^{n} b_n \) converges if \( b_n \) is decreasing and \( \lim_{n \to \infty} b_n = 0 \). Here, \( b_n = \frac{1}{5n^{1.1}} \).First, test if \( b_n \) is decreasing by considering the function \( f(n) = \frac{1}{5n^{1.1}} \). Since \( f(n) \) is a continuous, positive, and decreasing function for \( n \geq 1 \), and \( \lim_{n \to \infty} f(n) = 0 \), the series converges by the Alternating Series Test.
03

Test for Absolute Convergence

To check for absolute convergence, consider the series without the alternating factor: \( \sum_{n=1}^{\infty} \frac{1}{5n^{1.1}} \). This is a p-series with \( p = 1.1 \). A p-series \( \sum \frac{1}{n^p} \) converges if \( p > 1 \). Since 1.1 > 1, the series \( \sum_{n=1}^{\infty} \frac{1}{5n^{1.1}} \) converges.
04

Conclude the Type of Convergence

Since the series converges absolutely, it is also absolutely convergent. If a series converges absolutely, it also converges conditionally as a consequence.

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Key Concepts

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

Alternating Series
Alternating series are a fundamental concept in the study of series convergence. These series have terms that change sign with every subsequent term, usually characterized by the factor (-1)^n. This sign-changing nature can often lead to convergence even when the corresponding non-alternating series diverges.
  • An alternating series takes the form \( \sum (-1)^{n} b_n \).
  • For example, in the series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{5 n^{1.1}} \), the \((-1)^{n+1}\) ensures that the signs alternate with each term.
An important tool to determine the convergence of alternating series is the Alternating Series Test.
Absolute Convergence
Absolute convergence is a stronger form of convergence. If a series is absolutely convergent, it means that not only does the series converge, but the series formed by taking the absolute values of the terms also converges.
  • To check for absolute convergence, we consider the series without any negative signs from the alternating factor.
  • For the series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{5 n^{1.1}} \), the series we consider for absolute convergence is \( \sum_{n=1}^{\infty} \frac{1}{5n^{1.1}} \).
Determining the absolute convergence often involves proving that the series without the sign changes still converges.
P-Series
P-series are a specific type of series that serve as a standard example when testing for convergence; they are very useful due to their straightforward nature. These series take the form \( \sum \frac{1}{n^p} \).
  • The convergence of a p-series depends solely on the value of \( p \).
  • It converges if \( p > 1 \) and diverges if \( p \leq 1 \).
In the original exercise, we encounter a p-series when testing for absolute convergence: \( \sum_{n=1}^{\infty} \frac{1}{5n^{1.1}} \). Since \( p = 1.1 \) and 1.1 is greater than 1, this series converges.
Convergence Tests
Convergence tests are essential tools in determining whether a series converges or diverges. Different types of series might require different tests to determine their behavior.
  • Alternating Series Test: This test checks if an alternating series converges by evaluating whether the terms decrease and approach 0 as \( n \) goes to infinity.
  • Absolute Convergence Test: Determine convergence by considering the series formed by the absolute values of the terms.
  • P-Series Test: A critical test for p-series, based on the power \( p \). It informs us about convergence based solely on the parameter \( p \).
These tests are often combined to interpret the behavior of complex series correctly. Applying these especially helps to see if an alternating series can still be convergent by evaluating it without the alternating sign.

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Most popular questions from this chapter

Find the terms through \(x^{5}\) in the Maclaurin series for \(f(x) .\) Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x) .\) \(f(x)=\left(1-x^{2}\right)^{2 / 3}\)

Plot on the same axes the given function along with the Maclaurin polynomials of orders \(1,2,3\), and \(4 .\) $$ \sin e^{x} $$

Determine convergence or divergence for each of the series. Indicate the test you use. $$ \sum_{n=2}^{\infty}\left(1-\frac{1}{n}\right)^{n} $$

Many drugs are eliminated from the body in an exponential manner. Thus, if a drug is given in dosages of size \(C\) at time intervals of length \(t\), the amount \(A_{n}\) of the drug in the body just after the \((n+1)\) st dose is $$ A_{n}=C+C e^{-k t}+C e^{-2 k t}+\cdots+C e^{-n k t} $$ where \(k\) is a positive constant that depends on the type of drug. (a) Derive a formula for \(A\), the amount of drug in the body just after a dose, if a person has been on the drug for a very long time (assume an infinitely long time). (b) Evaluate \(A\) if it is known that one-half of a dose is eliminated from the body in 6 hours and doses of size 2 milligrams are given every 12 hours.

Suppose that \(\lim _{n \rightarrow \infty} n a_{n}=1 .\) Prove that \(\sum a_{n}\) diverges.
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