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Use the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals (see Problem 16 ). $$\sum_{n=1}^{\infty} \frac{n}{n^{2}+4}$$

Short Answer

Expert verified
The series diverges.

Step by step solution

01

Identify the Series

We are given the series \(\sum_{n=1}^{\infty} \frac{n}{n^{2}+4}\).We need to determine if this series converges or diverges using the integral test.
02

Define the Function for the Integral Test

Define the function \( f(x) = \frac{x}{x^{2}+4} \).We will use this function to apply the integral test.
03

Set Up the Improper Integral

Set up the improper integral \( \int_{1}^{\infty} \frac{x}{x^{2}+4} dx \).We need to evaluate this integral to apply the integral test.
04

Use a Substitution

Let \( u = x^{2} + 4 \). Then, \( du = 2x dx \), so \frac{1}{2} du = x dx\.Change the limits accordingly: when \(x = 1\), \(u = 5\); when \(x = \infty\), \u = \infty\.The integral becomes \(\int_{5}^{\infty} \frac{1}{2} \frac{1}{u} du}\).
05

Evaluate the Integral

The integral \(\int \frac{1}{2u} du \) evaluates to \(\frac{1}{2} \ln|u| \).So, the integral from \( 1 \text{ to } \text{infinity} \) becomes \(\frac{1}{2} [ \ln |\infty| - \ln |5| ] \).Since \( \ln|\infty| \) diverges to infinity, the overall result diverges.
06

Conclude Using the Integral Test

Since the improper integral \(\int_{1}^{\infty} \frac{x}{x^{2}+4} dx \) diverges, by the integral test, the series \(\sum_{n=1}^{\infty} \frac{n}{n^{2}+4} \) also diverges.

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Key Concepts

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

Series Convergence
In mathematics, understanding whether series converge or diverge is crucial. A series is essentially the sum of the terms of a sequence. When we talk about convergence, we mean that as the number of terms increases indefinitely, the sum approaches a specific finite value.
If, however, the sum grows without bound, we say that the series diverges.
There are different tests to determine series convergence, and in this exercise, we use the *integral test*.
Improper Integral
An improper integral is an integral with one or more infinite limits or an integrand that becomes infinite within the integration limits.
Evaluating improper integrals is essential in many concepts, including the integral test for series. For example, the integral \( \int_{1}^{\infty} \frac{x}{x^{2}+4} \) is improper because it extends to infinity.

To solve it, we need to convert it into a manageable form and evaluate it carefully.
In this scenario, we do not use lower limits when setting up our integral.
Substitution Method
The substitution method is a common technique used to simplify integrals. By changing variables, we can make an integral easier to solve.
For instance, in the exercise above, let’s say we choose \( u=x^{2}+4 \). Then, the differential \( du=2x \, dx \) allows us to rewrite the integral in simpler terms.
After substituting and changing the bounds, the original integral simplifies as follows:
\[ \int_{1}^{\infty} \frac{x}{x^{2}+4} \, dx \rightarrow \frac{1}{2} \int_{5}^{\infty} \frac{1}{u} \, du \].
This process helps in ultimately evaluating whether the original series converges or diverges.
Divergence
Determining the divergence of a series is as critical as identifying convergence. If a series diverges, it indicates that adding up all its terms results in an infinite value.
In the given problem, we find that the series \( \sum_{n=1}^{\infty} \frac{n}{n^{2}+4} \) diverges because the corresponding improper integral \[ \frac{1}{2} \left( \ln|\infty| - \ln|5| \right) \] diverges.
Since the logarithm of infinity is infinite, the integral does not settle to a finite value. Hence, the divergence of the integral confirms the divergence of the series.

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Most popular questions from this chapter

By computer, find a numerical approximation for the sum of each of the following series. $$\text { (a) } \sum_{n=1}^{\infty} \frac{n}{\left(n^{2}+1\right)^{2}}$$ $$\text { (b) } \sum_{n=2}^{\infty} \frac{\ln n}{n^{2}}$$ $$\text { (c) } \sum_{n=1}^{\infty} \frac{1}{n^{n}}$$

Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case. $$\sum_{n=1}^{\infty} \frac{(-1)^{n}(x+1)^{n}}{n}$$

Use the special comparison test to find whether the following series converge or diverge. $$\sum_{n=3}^{\infty} \frac{(n-\ln n)^{2}}{5 n^{4}-3 n^{2}+1}$$

(a) Using computer or tables (or see Chapter 7 , Section 11 ), verify that \(\sum_{n=1}^{\infty} 1 / n^{2}=\) \(\pi^{2} / 6=1.6449+,\) and also verify that the error in approximating the sum of the series by the first five terms is approximately 0.1813 . (b) \(\quad\) By computer or tables verify that \(\sum_{n=1}^{\infty}\left(1 / n^{2}\right)(1 / 2)^{n}=\pi^{2} / 12-(1 / 2)(\ln 2)^{2}=\) \(0.5822+,\) and that the sum of the first five terms is \(0.5815+.\) (c) Prove theorem (14.4). Hint: The error is \(\left|\sum_{N+1}^{\infty} a_{n} x^{n}\right| .\) Use the fact that the absolute value of a sum is less than or equal to the sum of the absolute values. Then use the fact that \(\left|a_{n+1}\right| \leq\left|a_{n}\right|\) to replace all \(a_{n}\) by \(a_{N+1},\) and write the appropriate inequality. Sum the geometric series to get the result.

Show that \(\sum_{n=2}^{\infty} 1 / n^{3 / 2}\) is convergent. What is wrong with the following "proof" that it diverges? $$\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{27}}+\frac{1}{\sqrt{64}}+\frac{1}{\sqrt{125}}+\cdots>\frac{1}{\sqrt{9}}+\frac{1}{\sqrt{36}}+\frac{1}{\sqrt{81}}+\frac{1}{\sqrt{144}}+\cdots$$ which is $$\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}+\dots=\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\right)$$. since the harmonic series diverges, the original series diverges. Hint: Compare \(3 n\) and \(n \sqrt{n}\).

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