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Chapter 11: Problem 25

Determine whether the series converges or diverges. $$ \sum_{n=1}^{\infty} \frac{e^{n}+1}{n e^{n}+1} $$

Short Answer

Expert verified
The series diverges.

Step by step solution

01

Identifying the Type of Series

The given series is \( \sum_{n=1}^{\infty} \frac{e^{n}+1}{n e^{n}+1} \). This series is not immediately recognized as a standard series like geometric or p-series. Therefore, we must use a test to determine convergence or divergence.
02

Choosing an Appropriate Test

Since the terms \( a_n = \frac{e^{n}+1}{n e^{n}+1} \) involve exponential functions, we will use the Limit Comparison Test to compare it with another known convergent or divergent series, such as \( \sum_{n=1}^{\infty} \frac{1}{n} \), a known divergent harmonic series.
03

Applying the Limit Comparison Test

We compare with \( b_n = \frac{1}{n} \). Calculate the limit: \[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{e^{n}+1}{n e^{n}+1}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n(e^n + 1)}{n e^n + 1} \]Simplifying gives:\[ \lim_{n \to \infty} \frac{e^n + 1}{e^n + \frac{1}{n}} = \lim_{n \to \infty} \frac{e^n(1 + \frac{1}{e^n})}{e^n(1 + \frac{1}{ne^n})} \]The terms \(\frac{1}{e^n}\) and \(\frac{1}{ne^n}\) approach zero as \(n\to\infty\). Thus, this simplifies to:\[ \lim_{n \to \infty} \frac{1 + 0}{1 + 0} = 1 \]
04

Interpreting the Limit Comparison Test Result

Since the limit here is 1, and \( \sum_{n=1}^{\infty} \frac{1}{n} \) is a divergent series, the Limit Comparison Test tells us that \( \sum_{n=1}^{\infty} \frac{e^{n}+1}{n e^{n}+1} \) also diverges.

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Key Concepts

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

Limit Comparison Test
The Limit Comparison Test is a tool used to determine whether a series converges or diverges by comparing it to another series with known behavior. Here's how it works:
  • Select a series with terms denoted as \( b_n \), which is known to converge or diverge.
  • Find the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \), where \( a_n \) represents the terms of the series in question.
  • If this limit is a positive finite number, both series converge or diverge together.
This test is particularly useful when dealing with series involving complex expressions such as exponential functions. By comparing with the simple \( \frac{1}{n} \) series (harmonic series), we can simplify our analysis.
Harmonic Series
The Harmonic Series is one of the simplest examples of a divergent series. It is represented as \(\sum_{n=1}^{\infty} \frac{1}{n}\).This series is well-known for its divergence, meaning it does not have a finite sum.
  • The partial sums of the harmonic series increase without bound as you add more terms.
  • Despite each term getting smaller, the sum grows indefinitely as \( n \) approaches infinity.
Understanding the harmonic series is crucial for recognizing other divergent behaviors using comparison tests, as it sets a benchmark for divergence common in calculus and analysis.
Divergent Series
A divergent series is a series that does not converge to a finite limit. Instead, as you sum more and more terms, the total grows without bound or oscillates indefinitely.
  • Unlike convergent series, divergent series do not approach a specific finite sum.
  • Examples include the harmonic series and many series involving exponential functions or improper fractions.
Identifying divergence is key in many mathematical fields, such as understanding the behaviors of sequences and series. This often involves tests like the Limit Comparison Test to determine how more complex series behave relative to simpler known divergent series.
Exponential Functions
Exponential functions, such as \( e^n \), play a vital role in many mathematical analyses, mainly due to their distinct growth properties:
  • They grow very rapidly compared to polynomial functions.
  • This rapid growth can often lead to divergence in series where exponential terms are present.
In the given exercise, the series \( \sum_{n=1}^{\infty} \frac{e^{n}+1}{n e^{n}+1} \) features exponential terms in both the numerator and the denominator. When applying convergence tests, understanding how exponential functions behave is essential, as they often dictate the series' overall convergence or divergence.

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Most popular questions from this chapter

Use the power series for \(\tan ^{-1} x\) to prove the following expression for \(\pi\) as the sum of an infinite series: $$ \pi=2 \sqrt{3} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) 3^{n}} $$

Use series to evaluate the limit. $$ \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1-\frac{1}{2} x}{x^{2}} $$

Use series to approximate the definite integral to within the indicated accuracy. $$ \int_{0}^{0.5} x^{2} e^{-x^{2}} d x \quad(| \text { error } |<0.001) $$

(a) Find the Taylor polynomials up to degree 5 for \(f(x)=\sin x\) centered at \(a=0 .\) Graph \(f\) and these polynomials on a common screen. (b) Evaluate \(f\) and these polynomials at \(x=\pi / 4, \pi / 2,\) and \(\pi\). (c) Comment on how the Taylor polynomials converge to \(f(x) .\)

The resistivity \(\rho\) of a conducting wire is the reciprocal of the conductivity and is measured in units of ohm-meters \((\Omega-\mathrm{m}) .\) The resistivity of a given metal depends on the temperature according to the equation \(\rho(t)=\rho_{20} e^{\alpha(t-2 \omega)}\) where \(t\) is the temperature in \(^{\circ} \mathrm{C}\). There are tables that list the values of \(\alpha\) (called the temperature coefficient) and \(\rho_{20}\) (the resistivity at \(20^{\circ} \mathrm{C}\) ) for various metals. Except at very low temperatures, the resistivity varies almost linearly with temperature and so it is common to approximate the expression for \(\rho(t)\) by its first-or second-degree Taylor polynomial at \(t=20 .\) $$ \begin{array}{l}{\text { (a) Find expressions for these linear and quadratic }} \\ {\text { approximations. }} \\ {\text { (b) For copper, the tables give } \alpha=0.0039 /^{\circ} \mathrm{C} \text { and }} \\ {\rho_{20}=1.7 \times 10^{-8} \Omega \text { -m. Graph the resistivity of copper }} \\\ {\text { and the linear and quadratic approximations for }} \\ {-250^{\circ} \mathrm{C} \leqslant t \leqslant 1000^{\circ} \mathrm{C} \text { . }} \\\ {\text { (c) For what values of } t \text { does the linear approximation agree }} \\ {\text { with the exponential expression to within one percent? }}\end{array} $$
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