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Chapter 9: Problem 7

Converge or diverge. Be sure to check that the conditions of the Integral Test are satisfied. $$\sum_{n=1}^{\infty} \frac{n}{n^{2}+4}$$

Short Answer

Expert verified
The series diverges by the Integral Test.

Step by step solution

01

Define the Function

To apply the Integral Test, we start by defining the continuous function corresponding to our series: \( f(x) = \frac{x}{x^2 + 4} \) for \( x \geq 1 \). This function is positive and continuous for all \( x \geq 1 \).
02

Check Decreasing Condition

Next, we need to check if \( f(x) = \frac{x}{x^2 + 4} \) is decreasing. We do this by finding its derivative \( f'(x) \). Using the quotient rule, we have:\[ f'(x) = \frac{(x^2 + 4)(1) - x(2x)}{(x^2 + 4)^2} = \frac{-x^2 + 4}{(x^2 + 4)^2} \]For \( x \geq 1 \), the numerator \(-x^2 + 4\) is negative, hence \( f(x) \) is decreasing.
03

Apply the Integral Test

Since \( f(x) = \frac{x}{x^2 + 4} \) is positive, continuous, and decreasing for \( x \geq 1 \), we can apply the Integral Test. We evaluate the improper integral:\[ \int_1^\infty \frac{x}{x^2 + 4} \, dx \]
04

Evaluate the Integral

Perform a substitution to evaluate the integral. Let \( u = x^2 + 4 \), then \( du = 2x \, dx \), so \( x \, dx = \frac{1}{2} du \). The limits change from \( x = 1 \) to \( u = 1^2 + 4 = 5 \), and \( u \to \infty \) as \( x \to \infty \). The integral becomes:\[ \frac{1}{2} \int_5^\infty \frac{1}{u} \, du \]Evaluating this, we have:\[ \frac{1}{2} \left[ \ln|u| \right]_5^\infty = \frac{1}{2} (\infty - \ln(5)) = \infty \]
05

Conclusion from the Integral Test

Since the 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.

Convergence
Convergence is a key concept in determining whether an infinite series, like \( \sum_{n=1}^{\infty} a_n \), adds up to a finite number. It occurs when the series approaches a specific value as more terms are added, meaning the partial sums of the series get closer to this finite limit. In practice, convergence is essential in calculus and analysis because it helps us understand the behavior and summability of series, which might represent functions or real-world phenomena. When a series converges, it means that no matter how many more terms are added, the additions become insignificant, consolidating into a single finite sum.The Integral Test is a handy tool for checking convergence, applying conditions like positivity, continuity, and decreasing nature over the series' infinite domain.
Divergence
When a series diverges, it means that, unlike convergent series, the partial sums do not settle to a finite limit. The series can grow infinitely large, oscillate, or fail to approach any particular value as more terms are added.In the exercise, the series \( \sum_{n=1}^{\infty} \frac{n}{n^2 + 4} \) was tested for divergence using the Integral Test. Since the improper integral \( \int_1^\infty \frac{x}{x^2 + 4} \, dx \) evaluated to infinity, the series diverges.Understanding divergence is crucial because sometimes, though the terms of the series seem to get smaller, the series as a whole doesn't sum to a limit. Evaluating divergence helps in recognizing series that don't have finite sums, important in preventing incorrect assumptions in calculus problems.
Improper Integrals
Improper integrals come into play when dealing with limits of integration at infinity or when an integrand becomes unbounded within the limits. Such integrals test the convergence or divergence of functions extended over infinite domains or at singular points. To compute an improper integral like \( \int_1^\infty \frac{x}{x^2 + 4} \, dx \), we often use substitution techniques to simplify the integration process. In the exercise, the substitution \( u = x^2 + 4 \) transformed the integral into a simpler form that could be integrated over \( u \) from 5 to infinity.The outcome revealed a divergent integral (approaching infinity) indicating that the corresponding series \( \sum_{n=1}^{\infty} \frac{n}{n^2 + 4} \) does not converge. Understanding improper integrals is key in many areas of mathematics and science, as they help handle infinite limits and discontinuities.

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Most popular questions from this chapter

Pythagorean triples \(\quad\) A triple of positive integers \(a, b,\) and \(c\) is called a Pythagorean triple if \(a^{2}+b^{2}=c^{2} .\) Let \(a\) be an odd positive integer and let $$b=\left\lfloor\frac{a^{2}}{2}\right\rfloor \quad \text { and } \quad c=\left\lceil\frac{a^{2}}{2}\right\rceil$$ be, respectively, the integer floor and ceiling for \(a^{2} / 2\). a. Show that \(a^{2}+b^{2}=c^{2} .\) (Hint: Let \(a=2 n+1\) and express \(b \text { and } c \text { in terms of } n .)\) b. By direct calculation, or by appealing to the accompanying figure, find $$\lim _{a \rightarrow \infty} \frac{\left\lfloor\frac{a^{2}}{2}\right\rfloor}{\left\lceil\frac{a^{2}}{2}\right\rceil}.$$

Prove that if \(\left\\{a_{n}\right\\}\) is a convergent sequence, then to every positive number \(\epsilon\) there corresponds an integer \(N\) such that for all \(m\) and \(n\), $$m>N \quad \text { and } \quad n>N \Rightarrow \quad\left|a_{m}-a_{n}\right|<\epsilon.$$

Sequences generated by Newton's method \(\quad\) Newton's method, applied to a differentiable function \(f(x),\) begins with a starting value \(x_{0}\) and constructs from it a sequence of numbers \(\left\\{x_{n}\right\\}\) that under favorable circumstances converges to a zero of \(f .\) The recursion formula for the sequence is $$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$ a. Show that the recursion formula for \(f(x)=x^{2}-a, a>0\) can be written as \(x_{n+1}=\left(x_{n}+a / x_{n}\right) / 2\). b. Starting with \(x_{0}=1\) and \(a=3,\) calculate successive terms of the sequence until the display begins to repeat. What number is being approximated? Explain.

Which series converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$\sum_{n=0}^{\infty} \frac{e^{n \pi}}{\pi^{n e}}$$

Which series converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$\sum_{n=0}^{\infty}\left(\frac{e}{\pi}\right)^{n}$$
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