/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Which of the series Converge abs... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The series converges conditionally by the Alternating Series Test.

Step by step solution

01

Analyze the given series

Consider the series \( \sum_{n=1}^{\infty} (-1)^n \frac{1}{\sqrt{n}} \). This is an alternating series because the terms change sign based on \((-1)^n\). First, consider whether the series converges absolutely.
02

Check for absolute convergence

A series \( \sum_{n=1}^{\infty} a_n \) converges absolutely if \( \sum_{n=1}^{\infty} |a_n| \) converges. For the given series, consider \( \sum_{n=1}^{\infty} \left| (-1)^n \frac{1}{\sqrt{n}} \right| = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \). This series is a \( p \)-series with \( p = \frac{1}{2} \), which is known to diverge because \( p \leq 1 \). Thus, the series does not converge absolutely.
03

Apply the Alternating Series Test

For an alternating series \( \sum_{n=1}^{\infty} (-1)^n a_n \) to converge, the terms \( a_n \) must: 1) be decreasing, 2) \( \lim_{n \to \infty} a_n = 0 \). Here \( a_n = \frac{1}{\sqrt{n}} \), which is decreasing for \( n \geq 1 \), and \( \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0 \). Therefore, the alternating 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.

Alternating series
An alternating series is a type of mathematical series where the signs of its terms alternate between positive and negative. The general form of such a series can be written as \( \sum_{n=1}^{\infty} (-1)^n a_n \), where the factor \((-1)^n\) or \((-1)^{n+1}\) ensures the alternation. For an alternating series to converge, two important conditions must be met:
  • The absolute values of the terms \( a_n \) should be decreasing, which means each term is smaller than the previous term.
  • The limit of \( a_n \) as \( n \) approaches infinity must be zero, \( \lim_{n \to \infty} a_n = 0 \).
In our example, the series \( \sum_{n=1}^{\infty} (-1)^n \frac{1}{\sqrt{n}} \) is indeed alternating, and since \( \frac{1}{\sqrt{n}} \) decreases as \( n \) increases and also tends to zero, this series converges by the Alternating Series Test.
Absolute convergence
Absolute convergence is a stronger form of convergence, implying that even when we take the absolute values of the terms in a series, the series should still converge. This is in contrast to conditional convergence, where only the original series converges. If a series \( \sum_{n=1}^{\infty} a_n \) converges absolutely, it means that the sum of the series formed by taking the absolute values of its terms, \( \sum_{n=1}^{\infty} |a_n| \), also converges.
To determine absolute convergence for the series \( \sum_{n=1}^{\infty} (-1)^n \frac{1}{\sqrt{n}} \), we consider \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \). Since this new series does not converge, as it is a \( p \)-series with \( p = \frac{1}{2} \leq 1 \), the original series does not converge absolutely.
Divergence
Divergence occurs when a series does not settle down to a single value as we sum its terms. In the context of series, if neither the limit of the sequence of partial sums exist nor does the absolute value series converge, the series is said to diverge. This means the sum grows without bound or oscillates without approaching a fixed sum.
In the case of the series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \), it diverges because it is a \( p \)-series with \( p = \frac{1}{2} \), which is known to diverge for any \( p \leq 1 \). Therefore, when analyzing the absolute convergence of our alternating series, the absolute series \( \sum_{n=1}^{\infty} \left| (-1)^n \frac{1}{\sqrt{n}} \right| = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \) does not meet the criteria for convergence, leading to divergence.
P-series
A \( p \)-series is a form of mathematical series defined by the formula \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). The behavior of a \( p \)-series largely depends on the value of \( p \). If \( p > 1 \), the series converges, and if \( p \leq 1 \), the series diverges.
For example, consider the series \( \sum_{n=1}^{\infty} \frac{1}{n^{1/2}} \), which is a \( p \)-series with \( p = \frac{1}{2} \). Since \( \frac{1}{2} < 1 \), this particular \( p \)-series diverges. This phenomenon is critical when assessing the absolute convergence of other series, such as the alternating series mentioned earlier. Understanding the properties of \( p \)-series helps in quickly identifying whether certain series diverge or converge.

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Most popular questions from this chapter

Taylor series for even functions and odd functions (Continuation of Section \(10.7,\) Exercise \(59 .\) ) Suppose that \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) converges for all \(x\) in an open interval \((-R, R) .\) Show that \begin{equation} \begin{array}{l}{\text { a. If } f \text { is even, then } a_{1}=a_{3}=a_{5}=\dots=0, \text { i.e., the Taylor }} \\ {\text { series for } f \text { at } x=0 \text { contains only even powers of } x .} \\ {\text { b. If } f \text { is odd, then } a_{0}=a_{4}=a_{4}=\cdots=0, \text { i.e., the Taylor }} \\ {\text { series for } f \text { at } x=0 \text { contains only odd powers of } x .}\end{array} \end{equation}

In Exercises \(17-44\) , use any method to determine if the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \frac{n !}{10^{n}}$$

\begin{equation} \begin{array}{l}{\text { a. Use Taylor's formula with } n=2 \text { to find the quadratic }} \\ {\text { approximation of } f(x)=(1+x)^{k} \text { at } x=0(k \text { a constant) }} \\ {\text { b. If } k=3, \text { for approximately what values of } x \text { in the interval }} \\ {[0,1] \text { will the error in the quadratic approximation be less }} \\ {\text { than } 1 / 100 ?}\end{array} \end{equation}

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Improving approximations of \(\pi\) \begin{equation} \begin{array}{l}{\text { a. Let } P \text { be an approximation of } \pi \text { accurate to } n \text { decimals. Show }} \\ {\text { that } P+\sin P \text { gives an approximation correct to } 3 n \text { decimals. }} \\ {\text { (Hint: Let } P=\pi+x . )} \\ {\text { b. Try it with a calculator. }}\end{array} \end{equation}

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