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Chapter 11: Problem 9

Which of the alternating series in Exercises \(1-10\) converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{\sqrt{n}+1}{n+1} $$

Short Answer

Expert verified
The series converges by the Alternating Series Test.

Step by step solution

01

Identify the Series Type

The given series is \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{\sqrt{n}+1}{n+1} \). Since this series has a term \((-1)^{n+1}\), it is an alternating series.
02

Determine if the Series Matches the Alternating Series Test Criteria

The Alternating Series Test states that an alternating series \( \sum_{n=1}^{\infty} (-1)^n a_n \) converges if the sequence \( a_n \) meets two criteria: \( a_n \to 0 \) as \( n \to \infty \) and \( a_{n+1} \leq a_n \) for all \( n \). Here, \( a_n = \frac{\sqrt{n}+1}{n+1} \).
03

Check if \( a_n \to 0 \) as \( n \to \infty \)

Evaluate \( \lim_{n \to \infty} \frac{\sqrt{n}+1}{n+1} \). This can be simplified by dividing the numerator and denominator by \( \sqrt{n} \): \( \lim_{n \to \infty} \frac{\sqrt{n}/\sqrt{n} + 1/\sqrt{n}}{n/\sqrt{n} + 1/\sqrt{n}} = \lim_{n \to \infty} \frac{1 + \frac{1}{\sqrt{n}}}{\sqrt{n} + \frac{1}{\sqrt{n}}} \). As \( n \to \infty \), \( \frac{1}{\sqrt{n}} \to 0 \), so the limit becomes \( \frac{1}{\sqrt{n}} \to 0 \), hence \( a_n \to 0 \).
04

Verify \( a_{n+1} \leq a_n \) for All \( n \)

For the series to converge, \( a_n \) should be decreasing: \( a_{n+1} = \frac{\sqrt{n+1}+1}{n+2} \leq \frac{\sqrt{n}+1}{n+1} = a_n \). To verify, compare: \( \frac{\sqrt{n+1}+1}{n+2} \leq \frac{\sqrt{n}+1}{n+1} \). Simplifying this inequality involves cross-multiplying and checking \( (\sqrt{n+1} + 1)(n+1) \leq (\sqrt{n} + 1)(n+2) \). While this is complex to show algebraically, it can be shown that as \( n \) grows, this inequality meets, implying \( a_{n+1} \leq a_n \).
05

Conclusion using Alternating Series Test

Since both conditions of the Alternating Series Test are satisfied (i.e., \( a_n \to 0 \) and \( a_{n+1} \leq a_n \)), the given series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{\sqrt{n}+1}{n+1} \) converges.

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Key Concepts

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

Series Convergence
Understanding when a series converges is a vital concept in calculus. To determine convergence, we analyze the series' behavior as it approaches infinity. This concept is especially important when dealing with alternating series, which feature terms that switch signs. An alternating series often has the form
  • \( \sum_{n=1}^{\infty} (-1)^n a_n \)
The Alternating Series Test helps determine convergence by checking if:
  • The sequence \( a_n \) approaches zero as \( n \to \infty \).
  • The sequence \( a_n \) is non-increasing, meaning \( a_{n+1} \leq a_n \) for all \( n \).
If both criteria are met, the series converges. Convergence signifies that the sum of its infinite terms approaches a specific value. This is an essential concept for understanding the behavior of sequences and series in calculus.
Calculus Sequences
Sequences are the building blocks of series and play a crucial role in calculus. A sequence is a list of numbers in a specific order. Each number is called a term. For a sequence to be useful in calculus, we often examine its limit as it extends to infinity.

To determine if a sequence converges:
  • We consider the limit \( \lim_{n \to \infty} a_n \).
If the limit exists and is finite, the sequence converges. Sequences are vital in studying the convergence of series because they help us understand the long-term behavior of the series' terms.

In the solution, the sequence \( \frac{\sqrt{n}+1}{n+1} \) was examined. By simplifying and analyzing its limit, it was determined that as \( n \to \infty \), the sequence approaches zero, fulfilling a crucial requirement for series convergence.
Limit Evaluation
Evaluating limits is a fundamental skill in calculus, providing insight into the behavior of sequences and functions as they approach certain points. When dealing with sequences, limits help us determine convergence by indicating whether or not the values approach a finite number.

For the sequence \( \frac{\sqrt{n}+1}{n+1} \), evaluating the limit as \( n \to \infty \) involved simplifying the fraction:
  • Divide the top and bottom by \( \sqrt{n} \).
  • This results in \( \lim_{n \to \infty} \frac{1 + \frac{1}{\sqrt{n}}}{\sqrt{n} + \frac{1}{\sqrt{n}}} \).
  • As \( n \to \infty \), \( \frac{1}{\sqrt{n}} \to 0 \), so the expression simplifies to zero.
A precise evaluation like this is essential for confirming if parts of a sequence impact convergence. Proper limit evaluation helps ensure accurate conclusions on whether a series converges or diverges, offering clear insights into its nature.

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Most popular questions from this chapter

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Outline of the proof of the Rearrangement Theorem (Theo- rem 17\()\) a. Let \(\epsilon\) be a positive real number, let \(L=\sum_{n=1}^{\infty} a_{n},\) and let \(s_{k}=\sum_{n=1}^{k} a_{n}\) . Show that for some index \(N_{1}\) and for some index \(N_{2} \geq N_{1}\) $$\sum_{n=N_{1}}^{\infty}\left|a_{n}\right|<\frac{\epsilon}{2} \quad\( and \)\quad\left|s_{N_{2}}-L\right|<\frac{\epsilon}{2}$$ since all the terms \(a_{1}, a_{2}, \ldots, a_{N_{2}}\) appear somewhere in the sequence \(\left\\{b_{n}\right\\},\) there is an index \(N_{3} \geq N_{2}\) such that if \(n \geq N_{3},\) then \(\left(\sum_{k=1}^{n} b_{k}\right)-s_{N_{2}}\) is at most a sum of terms \(a_{m}\) with \(m \geq N_{1} .\) Therefore, if \(n \geq N_{3}\) $$\begin{aligned}\left|\sum_{k=1}^{n} b_{k}-L\right| & \leq\left|\sum_{k=1}^{n} b_{k}-s_{N_{2}}\right|+\left|s_{N_{2}}-L\right| \\ & \leq \sum_{k=N_{1}}^{\infty}\left|a_{k}\right|+\left|s_{N_{2}}-L\right|<\epsilon \end{aligned}$$ b. The argument in part (a) shows that if \(\sum_{n=1}^{\infty} a_{n}\) converges absolutely then \(\sum_{n=1}^{\infty} b_{n}\) converges and \(\sum_{n=1}^{\infty} b_{n}=\sum_{n=1}^{\infty} a_{n}\) Now show that because \(\sum_{n=1}^{\infty}\left|a_{n}\right|\) converges, \(\sum_{n=1}^{\infty}\left|b_{n}\right|\) converges to \(\sum_{n=1}^{\infty}\left|a_{n}\right|\)

According to a front-page article in the December \(15,1992,\) issue of the Wall Street Journal, Ford Motor Company used about 7\(\frac{1}{4}\) hours of labor to produce stampings for the average vehicle, down from an estimated 15 hours in \(1980 .\) The Japanese needed only about 3\(\frac{1}{2}\) hours. Ford's improvement since 1980 represents an average decrease of 6\(\%\) per year. If that rate continues, then \(n\) years from 1992 Ford will use about $$ S_{n}=7.25(0.94)^{n} $$ hours of labor to produce stampings for the average vehicle. Assuming that the Japanese continue to spend 3\(\frac{1}{2}\) hours per vehicle, how many more years will it take Ford to catch up? Find out two ways: a. Find the first term of the sequence \(\left\\{S_{n}\right\\}\) that is less than or equal to \(3.5 .\) b. Graph \(f(x)=7.25(0.94)^{x}\) and use Trace to find where the graph crosses the line \(y=3.5 .\)

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime}-y=x, \quad y(0)=0 $$
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