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Chapter 11: Problem 8

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

Short Answer

Expert verified
The series converges by the Alternating Series Test.

Step by step solution

01

Identify the Series

The series given is \( \sum_{n=1}^{\infty}(-1)^{n} \ln \left(1+\frac{1}{n}\right) \). It is an alternating series since it has the term \((-1)^n\), which causes the sign to alternate for each consecutive term.
02

Apply the Alternating Series Test

The Alternating Series Test states that an alternating series \( \sum (-1)^n a_n \) converges if two conditions are met: 1. The absolute value of the terms \( a_n = \ln(1+\frac{1}{n}) \) decreases monotonically, \(a_{n+1} \leq a_n\). 2. The limit of \( a_n \) as \( n \to \infty \) is zero, \( \lim_{n \to \infty} a_n = 0 \).
03

Verify Decreasing Terms

Examine the term \( a_n = \ln(1+\frac{1}{n}) \). Since the natural logarithm \( \ln(x) \) increases with \( x \), as \( n \) increases, \( 1+\frac{1}{n} \) decreases. Thus, \( \ln(1+\frac{1}{n}) \) decreases with \( n \). Therefore, the first condition of monotonicity is satisfied.
04

Determine the Limit of Terms

Compute the limit: \[ \lim_{n \to \infty} \ln\left(1+\frac{1}{n}\right) \]As \( n \to \infty \), \( \frac{1}{n} \to 0 \), thus\( 1+\frac{1}{n} \to 1 \).Since \( \ln(1) = 0 \), we find\[ \lim_{n \to \infty} \ln\left(1+\frac{1}{n}\right) = 0. \] The second condition of the Alternating Series Test is satisfied.
05

Conclude Convergence

Since both conditions of the Alternating Series Test are satisfied, the series \( \sum_{n=1}^{\infty}(-1)^{n} \ln \left(1+\frac{1}{n}\right) \) converges.

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Key Concepts

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

Convergence
Convergence is a core idea when working with series in mathematics. In simple terms, a series is said to converge if the sequence of its partial sums tends to a limit as the number of terms goes to infinity. For an alternating series, which has terms that alternate in sign, the Alternating Series Test is often employed. This test has two vital conditions:
  • The absolute values of the terms in the series must decrease monotonically; that is, each term must be smaller than the one before it.
  • The limit of the absolute value of the terms, as they go to infinity, must be zero.
A series converging means it can be summed up to a finite number, even if it has infinitely many terms. This is an essential concept for evaluating whether the series makes sense within a mathematical or practical context. Understanding convergence helps in determining the behavior of series and ensuring calculations involving them are reliable.
Monotonicity
Monotonicity refers to the behavior of a sequence or series in terms of being consistently non-increasing or non-decreasing. When verifying a series for convergence using the Alternating Series Test, ensuring the terms decrease monotonically is crucial. In the context of the series \(\sum_{n=1}^{\infty}(-1)^{n} \ln \left(1+\frac{1}{n}\right)\), each term is \(\ln(1+\frac{1}{n})\), these terms decrease as \(n\) increases because:
  • The expression inside the logarithm, \(1+\frac{1}{n}\), becomes smaller as \(n\) grows.
  • The nature of the natural logarithm function is such that it increases with its input. Therefore, if the input decreases, the value of the logarithm also decreases.
Understanding monotonicity helps to establish that the sequence of terms neither oscillates unpredictably nor grows away, thus maintaining a necessary aspect of series convergence.
Natural Logarithm
The natural logarithm, denoted as \(\ln(x)\), is a logarithm to the base \(e\), where \(e\) is approximately 2.718. It is a fundamental concept in calculus and mathematical analysis due to its natural growth rate properties. In the study of alternating series, the behavior of natural logarithms can affect convergence.For the series \(\sum_{n=1}^{\infty}(-1)^{n} \ln \left(1+\frac{1}{n}\right)\), we observe:
  • The function \(\ln(x)\) is continuous and differentiable for \(x > 0\).
  • \(\ln(x)\) increases as \(x\) increases. This characteristic is used to determine monotonicity in our series.
The natural logarithm serves not only as a mathematical function but also bridges exponential growth models with linear perceptions, making it a universal tool in mathematical research and problem-solving. Understanding \(\ln(x)\) is necessary for accurately assessing series involving natural logarithms.

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Most popular questions from this chapter

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=\left(1+\frac{0.5}{n}\right)^{n} $$

Logistic difference equation The recursive relation $$ a_{n+1}=r a_{n}\left(1-a_{n}\right) $$ is called the logistic difference equation, and when the initial value \(a_{0}\) is given the equation defines the logistic sequence \(\left\\{a_{n}\right\\} .\) Throughout this exercise we choose \(a_{0}\) in the interval \(03.57\) . Choose \(r=3.65\) and calculate and plot the first 300 terms of \(\left\\{a_{n}\right\\} .\) Observe how the terms wander around in an unpredictable, chaotic fashion. You cannot predict the value of \(a_{n+1}\) from previous values of the sequence. g. For \(r=3.65\) choose two starting values of \(a_{0}\) that are close together, say, \(a_{0}=0.3\) and \(a_{0}=0.301 .\) Calculate and plot the first 300 values of the sequences determined by each starting value. Compare the behaviors observed in your plots. How far out do you go before the corresponding terms of your two sequences appear to depart from each other? Repeat the exploration for \(r=3.75 .\) Can you see how the plots look different depending on your choice of \(a_{0} ?\) We say that the logistic sequence is sensitive to the initial condition a_{0} .

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=123456^{1 / n} $$

In Exercises \(33-36\) , use series to estimate the integrals' values with an error of magnitude less than \(10^{-3} .\) (The answer section gives the integrals' values rounded to five decimal places.) $$ \int_{0}^{0.1} \frac{1}{\sqrt{1+x^{4}}} d x $$

For what values of \(a,\) if any, do the series in converge? $$\sum_{n=3}^{\infty}\left(\frac{1}{n-1}-\frac{2 a}{n+1}\right)$$
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