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Chapter 1: Problem 11

Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=4}^{\infty} \frac{2 n}{n^{2}-9}$$

Short Answer

Expert verified
The series diverges by the Limit Comparison Test.

Step by step solution

01

- Preliminary Test (Divergence Test)

First, apply the Divergence Test which states that if the limit of the sequence terms doesn't approach 0, the series diverges. Evaluate the limit of the general term as follows: \[ \text{Let } a_n = \frac{2n}{n^2 - 9} \] Calculate: \[ \text{lim}_{n \to \text{infty}} \frac{2n}{n^2 - 9} \] Divide the numerator and denominator by \( n \): \[ \text{lim}_{n \to \text{infty}} \frac{2}{n - \frac{9}{n}} \] As \( n \to \text{infty} \), \( \frac{9}{n} \to 0 \). Thus we have: \[ \text{lim}_{n \to \text{infty}} \frac{2}{n} = 0 \] Since the limit is 0, the Divergence Test is inconclusive.
02

- Choose a Convergence Test: Comparison Test

The given series resembles a rational function where the degree of polynomial in the denominator is greater than the degree in the numerator. The Comparison Test will be suitable. Compare the given series with a simpler series: \[ 0 < \frac{2n}{n^2 - 9} < \frac{2n}{n^2/2} = \frac{4}{n}, \forall n \text{ sufficiently large} \] Consider the p-series: \( \sum_{n=4}^{\text{infty}} \frac{1}{n} \) which diverges because \( p = 1 \leq 1 \).
03

- Application of the Limit Comparison Test

To rigorously apply the Comparison Test, use the Limit Comparison Test with \( b_n = \frac{1}{n} \): \[ \text{Let } a_n = \frac{2n}{n^2 - 9} \text{ and } b_n = \frac{1}{n} \] Compute the limit: \[ \text{lim}_{n \to \text{infty}} \frac{a_n}{b_n} = \text{lim}_{n \to \text{infty}} \frac{\frac{2n}{n^2 - 9}}{\frac{1}{n}} = \text{lim}_{n \to \text{infty}} \frac{2n^2}{n^2 - 9} = \] \[ \text{lim}_{n \to \text{infty}} \frac{2n^2}{n^2} = 2 \] Since the limit is a finite, nonzero positive number (\(2\)), by the Limit Comparison Test, \( \sum a_n \) and \( \sum b_n \) have the same behavior. Since \( \sum b_n = \sum \frac{1}{n} \) diverges, \( \sum_{n=4}^{\text{infty}} \frac{2n}{n^2 - 9} \) also diverges.

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Key Concepts

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

Divergence Test
When testing a series for convergence or divergence, it's wise to begin with the Divergence Test.
This test checks if the limit of the sequence's terms does not approach zero as n approaches infinity.
If the limit is not zero, the series diverges. For our series:
Let: \( a_n = \frac{2n}{n^2 - 9} \)
  • Calculate the limit: \( \text{lim}_{n \to \text{infty}} a_n \)
  • Divide each term by \( n \): \( \text{lim}_{n \to \text{infty}} \frac{2}{n - \frac{9}{n}} \)
  • As \( n \to \text{infty} \), \( \frac{9}{n} \) approaches 0, giving us: \( \text{lim}_{n \to \text{infty}} \frac{2}{n} = 0 \)
Since the limit is 0, the Divergence Test is inconclusive here.
Comparison Test
When the Divergence Test is inconclusive, other tests like the Comparison Test can help.
This test involves comparing the series to a known benchmark series. Choose: \( \frac{4}{n} \) because:
  • For large n, \( \frac{2n}{n^2 - 9} \) approximately behaves like \( \frac{4}{n} \)
  • Comparison: \( 0 < \frac{2n}{n^2 - 9} < \frac{4}{n} \)
Given the benchmark p-series: \( \frac{1}{n} \) diverges because exponent p = 1.
If the series being compared diverges, the original series also diverges.
Limit Comparison Test
For a stricter comparison, use the Limit Comparison Test. This test compares two series by computing the limit of their ratio.
If the limit is a finite, non-zero constant, both series either converge or diverge together.
Consider:
  • \( a_n = \frac{2n}{n^2 - 9} \)
  • \( b_n = \frac{1}{n} \)
Calculate the limit: \( \text{lim}_{n \to \text{infty}} \frac{a_n}{b_n} \)
  • Combine ratios: \( \text{lim}_{n \to \text{infty}} \frac{\frac{2n}{n^2 - 9}}{\frac{1}{n}} = \frac{2n^2}{n^2 - 9} = \frac{2n^2}{n^2} = 2 \)
The limit being 2 (non-zero, positive) means both series, \( \frac{2n}{n^2 - 9} \) and \( \frac{1}{n} \) have the same divergence behavior.
Since \( \frac{1}{n} \) diverges, so does \( \frac{2n}{n^2 - 9} \).
p-series
The p-series is a significant concept in understanding series convergence.
A p-series is of the form \( \frac{1}{n^p} \). Here's how convergence depends on p:
  • If \( p > 1 \), the series converges.
  • If \( p \le 1 \), the series diverges.
For example, the harmonic series \( \frac{1}{n} \) is a p-series where p = 1.
It diverges because p \( \le 1 \).
Recognizing and comparing against p-series helps determine the behavior of more complex series.

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Most popular questions from this chapter

Use Maclaurin series to evaluate each of the following. Although you could do them by computer, you can probably do them in your head faster than you can type them into the computer. So use these to practice quick and skillful use of basic series to make simple calculations. $$\lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{x^{2}}$$

Find the interval of convergence, including end-point tests: $$\sum_{n=1}^{\infty} \frac{(x+2)^{n}}{(-3)^{n} \sqrt{n}}$$

A tall tower of circular cross section is reinforced by horizontal circular disks (like large coins), one meter apart and of negligible thickness. The radius of the disk at height \(n\) is \(1 /(n \ln n)(n \geq 2)\) Assuming that the tower is of infinite height: (a) Will the total area of the disks be finite or not? Hint: Can you compare the series with a simpler one? (b) If the disks are strengthened by wires going around their circumferences like tires, will the total length of wire required be finite or not? (c) Explain why there is not a contradiction between your answers in (a) and (b). That is, how is it possible to start with a set of disks of finite area, remove a little strip around the circumference of each, and get an infinite total length of these strips? Hint: Think about units-you can't compare area and length. Consider two cases: (1) Make the width of each strip equal to one percent of the radius of the disk from which you cut it. Now the total length is infinite but what about the total area? (2) Try to make the strips all the same width; what happens? Also see Chapter 5, Problem 3.31(b).

Use Maclaurin series to evaluate each of the following. Although you could do them by computer, you can probably do them in your head faster than you can type them into the computer. So use these to practice quick and skillful use of basic series to make simple calculations. $$\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}$$

(a) It is clear that you (or your computer) can't find the sum of an infinite series just by adding up the terms one by one. For example, to get \(\zeta(1.1)=\) \(\sum_{n=1}^{\infty} 1 / n^{1.1}(\text { see Problem } 15.22)\) with error \(<0.005\) takes about \(10^{33}\) terms. To see a simple alternative (for a series of positive decreasing terms) look at Figures 6.1 and 6.2. Show that when you have summed \(N\) terms, the sum \(R_{N}\) of the rest of the series is between \(I_{N}=\int_{N}^{\infty} a_{n} d n\) and \(I_{N+1}=\int_{N+1}^{\infty} a_{n} d n\) (b) Find the integrals in (a) for the \(\zeta(1.1)\) series and verify the claimed number of terms needed for error \(<0.005 .\) Hint: Find \(N\) such that \(I_{N}=0.005 .\) Also find upper and lower bounds for \(\zeta(1.1)\) by computing \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N}^{\infty} n^{-1.1} d n\) and \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N+1}^{\infty} n^{-1.1} d n\) where \(N\) is far less than \(10^{33} .\) Hint: You want the difference between the upper and lower limits to be about \(0.005 ;\) find \(\mathrm{N}\) so that term \(a_{N}=0.005\).
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