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Which of the series in Exercises \(11-44\) converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1+n}{n^{2}} $$

Short Answer

Expert verified
The series converges conditionally but not absolutely.

Step by step solution

01

Determine Absolute Convergence

To assess absolute convergence, we examine the series of absolute terms:\[\sum_{n=1}^{\infty} \left|(-1)^{n+1} \frac{1+n}{n^2}\right| = \sum_{n=1}^{\infty} \frac{1+n}{n^2}.\]Simplifying the series:\[\frac{1+n}{n^2} = \frac{1}{n^2} + \frac{1}{n}.\]The series becomes:\[\sum_{n=1}^{\infty} \frac{1}{n^2} + \sum_{n=1}^{\infty} \frac{1}{n}.\]
02

Analyze Component Series

The first component, \(\sum_{n=1}^{\infty} \frac{1}{n^2}\), is a p-series with \(p=2\), which converges because \(p > 1\).The second component, \(\sum_{n=1}^{\infty} \frac{1}{n}\), is a harmonic series, which diverges.
03

Conclude Absolute Convergence

Since one of the component series, \(\sum_{n=1}^{\infty} \frac{1}{n}\), diverges, the entire series \(\sum_{n=1}^{\infty} \left|(-1)^{n+1} \frac{1+n}{n^2}\right|\) diverges.Therefore, the original series does not converge absolutely.
04

Use Alternating Series Test

We now check the original series \(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1+n}{n^2}\) using the Alternating Series Test.Let \(a_n = \frac{1+n}{n^2}\).1. Check that \(a_n\) is decreasing:\[a_{n} = \frac{1+n}{n^2}, \quad a_{n} - a_{n+1} \approx \frac{1}{n} - \frac{1}{n+1}.\]Since \( a_{n+1} < a_{n} \) for all sufficiently large \( n \), \(a_n\) is decreasing.2. Check if \(\lim_{n \to \infty} a_n = 0\):\[\lim_{n \to \infty}\frac{1+n}{n^2} = \lim_{n \to \infty}\left( \frac{1}{n^2} + \frac{1}{n}\right) = 0.\]Both conditions are met, so the series converges by the Alternating Series Test.

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Key Concepts

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

Absolute Convergence
When we discuss absolute convergence of a series, we're interested in whether the series formed by taking the absolute values of its terms converges. This consideration is key because it informs us not just if the series sums to a finite value, but if it does so without regard to the sign of its terms.
Imagine checking if \[\sum_{n=1}^{\infty} \left|(-1)^{n+1} \frac{1+n}{n^2}\right|\]converges. We aim to simplify it into more familiar series. Here, the series transforms into:
- \(\sum_{n=1}^{\infty} \frac{1}{n^2} + \sum_{n=1}^{\infty} \frac{1}{n}\)These are more recognizable series whose behavior we can analyze specifically. The absolute convergence check shows whether we're on safe ground treating the terms without their alternating sign.
Alternating Series Test
The Alternating Series Test is a lifeline when dealing with series that switch between positive and negative terms. It's named because the series alternates in sign, and we can use this test when two key conditions are met: the terms decrease in magnitude and approach zero.
For the series \[\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1+n}{n^2}\]notice its alternating nature thanks to \((-1)^{n+1}\) factor. The test checks are:
  • The sequence \(a_n = \frac{1 + n}{n^2}\) is decreasing.
  • The limit as \(n\) approaches infinity of \(a_n\) is zero.
If both conditions hold true, as they do here, the series converges. This test doesn't confirm absolute convergence but tells us that the series will still lead to a finite sum despite the lack of absolute convergence.
p-Series
A p-series is a type of mathematical series of the form:\[\sum_{n=1}^{\infty} \frac{1}{n^p}\]where \(p\) is a positive constant. The behavior of a p-series is contingent upon the value of \(p\):
  • If \(p > 1\), the series converges.
  • If \(p \leq 1\), the series diverges.
One of our component series is:\[\sum_{n=1}^{\infty} \frac{1}{n^2}\]This is a p-series with \(p = 2\), which surpasses 1, ensuring convergence. Recognizing and understanding p-series help predict behavior, especially within more complex expressions.
Harmonic Series
The harmonic series is a classic example in mathematical analysis which exhibits divergence, famously despite its terms getting smaller, the series:\[\sum_{n=1}^{\infty} \frac{1}{n}\]do not sum to a finite number. This might be surprising at first glance because one might expect that as you keep adding smaller and smaller values, the total should stabilize eventually.
Nevertheless, the harmonic series continues to grow without bound. Therefore, it diverges. This property helps explain why certain series do not converge absolutely — a feature directly tied to one of the components of our studied series. Even though the \(\frac{1}{n}\) term shrinks, over an infinite summation, it results in an infinite sum.

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Most popular questions from this chapter

Integrate the first three nonzero terms of the Taylor series for tan \(t\) from 0 to \(x\) to obtain the first three nonzero terms of the Taylor series for \(\ln \sec x .\)

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} .

Show that if \(\sum_{n=1}^{\infty} a_{n}\) and \(\sum_{n=1}^{\infty} b_{n}\) both converge absolutely, then so does $$ \begin{array}{ll}{\text { a. } \sum_{n=1}^{\infty}\left(a_{n}+b_{n}\right)} & {\text { b. } \sum_{n=1}^{\infty}\left(a_{n}-b_{n}\right)} \\ {\text { c. } \sum_{n=1}^{\infty} k a_{n} \quad(k \text { any number })}\end{array} $$

Prove that \(\lim _{n \rightarrow \infty} x^{1 / n}=1,(x>0)\).

Euler's constant Graphs like those in Figure 11.8 suggest that as \(n\) increases there is little change in the difference between the sum $$1+\frac{1}{2}+\cdots+\frac{1}{n}$$ and the integral $$\ln n=\int_{1}^{n} \frac{1}{x} d x$$ To explore this idea, carry out the following steps. a. By taking \(f(x)=1 / x\) in the proof of Theorem 9 , show that $$\ln (n+1) \leq 1+\frac{1}{2}+\cdots+\frac{1}{n} \leq 1+\ln n$$ or $$0<\ln (n+1)-\ln n \leq 1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \leq 1$$ Thus, the sequence $$ a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n $$ is bounded from below and from above. b. Show that $$ \frac{1}{n+1}<\int_{n}^{n+1} \frac{1}{x} d x=\ln (n+1)-\ln n $$ and use this result to show that the sequence \(\left\\{a_{n}\right\\}\) in part (a) is decreasing. since a decreasing sequence that is bounded from below converges (Exercise 107 in Section 11.1\()\) , the numbers \(a_{n}\) defined in part (a) converge: $$1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \rightarrow \gamma$$ The number \(\gamma,\) whose value is \(0.5772 \ldots,\) is called Euler's constant. In contrast to other special numbers like \(\pi\) and \(e,\) no other expression with a simple law of formulation has ever been found for \(\gamma .\)

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