/*! 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 27 Determine whether the following ... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the following series converge. $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{\sqrt{k^{2}+4}}$$

Short Answer

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Question: Determine whether the following series converges or not: $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{\sqrt{k^{2}+4}}$$ Answer: The series converges.

Step by step solution

01

Write down the sequence

We need to write down the terms of the sequence, which are given by: $$a_k = \frac{(-1)^{k}}{\sqrt{k^{2}+4}}$$
02

Check the terms are decreasing

To ensure the terms are decreasing, we check that \(a_{k+1} \leq a_k\) for all \(k\): $$\frac{(-1)^{k+1}}{\sqrt{(k+1)^{2}+4}} \leq \frac{(-1)^{k}}{\sqrt{k^{2}+4}}$$ Since the exponents of \(-1\) alternate signs, we'll consider the absolute value of the terms instead. $$\left|\frac{(-1)^{k+1}}{\sqrt{(k+1)^{2}+4}}\right| \leq \left|\frac{(-1)^{k}}{\sqrt{k^{2}+4}}\right|$$ $$\frac{1}{\sqrt{(k+1)^{2}+4}} \leq \frac{1}{\sqrt{k^{2}+4}}$$ Since denominators are positive, we can reverse the inequality if we reverse the fractions: $$\sqrt{(k+1)^{2}+4} \geq \sqrt{k^{2}+4}$$ This inequality holds true as \((k+1)^2 > k^2\). So, the terms are indeed decreasing.
03

Determine the limit

We need to find the limit of the sequence as \(k\) approaches infinity: $$\lim_{k\to\infty} \frac{(-1)^{k}}{\sqrt{k^{2}+4}}$$ We consider the absolute value of the sequence: $$\lim_{k\to\infty} \left|\frac{(-1)^{k}}{\sqrt{k^{2}+4}}\right| = \lim_{k\to\infty} \frac{1}{\sqrt{k^{2}+4}}$$ As \(k\) approaches infinity, the denominator grows without bound, and the limit evaluates to zero: $$\lim_{k\to\infty} \frac{1}{\sqrt{k^{2}+4}} = 0$$
04

Apply the Alternating Series Test

Since the terms of the sequence are decreasing and the limit of the sequence is zero, the Alternating Series Test tells us that the series converges: $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{\sqrt{k^{2}+4}}\ \text{converges}$$

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Key Concepts

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

Convergent Series
A convergent series is one where the sum of all its terms approaches a specific value. This is a fundamental concept in calculus and analysis. In simple terms, as you add more and more terms of the sequence, the total sum gets closer and closer to a particular number, rather than continuing to grow indefinitely or oscillate.
In the given exercise, we are looking at an alternating series, which switches between positive and negative terms. This is crucial because such series often have an easier path to convergence provided certain conditions are met.
  • For a series to converge, especially an alternating one, the absolute value of the terms should generally get smaller.
  • The terms should approach zero as you go further down the series.
Understanding how and why a series converges helps in many areas, from theoretical mathematics to real-world applications such as physics and engineering.
Sequence Limit
A sequence limit refers to the value that a sequence approaches as the index goes to infinity. It is a cornerstone of mathematical analysis as it helps define continuity, integrals, and derivatives. In this particular exercise, finding the limit of the sequence's terms as the number of terms goes towards infinity is a critical step.
In step 3 of the solution, we analyze the terms of the sequence:
  • The sequence is given as \( a_k = \frac{(-1)^{k}}{\sqrt{k^{2}+4}} \).
  • Taking the limit as \( k \to \infty \), we simplify to \( \lim_{k \to \infty} \frac{1}{\sqrt{k^{2}+4}} \).
  • As \( k \) increases, the denominator grows significantly, making the whole fraction approach zero.
This means as you pick larger and larger indices, the sequence's terms get very close to zero, contributing to the series' convergence.
Decreasing Sequence
A decreasing sequence is one where each term is less than or equal to the previous term. Recognizing a decreasing sequence is vital when applying convergence tests, like the Alternating Series Test. This step focuses on ensuring the condition for the decreasing nature of a sequence.
In the given problem, we checked whether the sequence \( a_k \) decreased:
  • Ignoring the alternating sign, we observe the absolute value: \( \left|\frac{(-1)^{k+1}}{\sqrt{(k+1)^{2}+4}}\right| \leq \left|\frac{(-1)^{k}}{\sqrt{k^{2}+4}}\right| \).
  • The inequality simplifies to checking \( \sqrt{(k+1)^2 + 4} \geq \sqrt{k^2 + 4} \).
  • This inequality holds true because \( (k+1)^2 > k^2 \), confirming a decreasing sequence when absolute values are considered.
Having a decreasing sequence of terms in magnitude is a key checkpoint for many convergence tests, ensuring the series has a clear path to sum up to a finite value.

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Most popular questions from this chapter

In Section \(8.3,\) we established that the geometric series \(\sum r^{k}\) converges provided \(|r| < 1\). Notice that if \(-1 < r<0,\) the geometric series is also an alternating series. Use the Alternating Series Test to show that for \(-1 < r <0\), the series \(\sum r^{k}\) converges.

Consider the following situations that generate a sequence. a. Write out the first five terms of the sequence. b. Find an explicit formula for the terms of the sequence. c. Find a recurrence relation that generates the sequence. d. Using a calculator or a graphing utility, estimate the limit of the sequence or state that it does not exist. The Consumer Price Index (the CPI is a measure of the U.S. cost of living) is given a base value of 100 in the year \(1984 .\) Assume the CPI has increased by an average of \(3 \%\) per year since \(1984 .\) Let \(c_{n}\) be the CPI \(n\) years after \(1984,\) where \(c_{0}=100\)

For a positive real number \(p,\) the tower of exponents \(p^{p^{p}}\) continues indefinitely and the expression is ambiguous. The tower could be built from the top as the limit of the sequence \(\left\\{p^{p},\left(p^{p}\right)^{p},\left(\left(p^{p}\right)^{p}\right)^{p}, \ldots .\right\\},\) in which case the sequence is defined recursively as \(a_{n+1}=a_{n}^{p}(\text { building from the top })\) where \(a_{1}=p^{p} .\) The tower could also be built from the bottom as the limit of the sequence \(\left\\{p^{p}, p^{\left(p^{p}\right)}, p^{\left(p^{(i)}\right)}, \ldots .\right\\},\) in which case the sequence is defined recursively as \(a_{n+1}=p^{a_{n}}(\text { building from the bottom })\) where again \(a_{1}=p^{p}\). a. Estimate the value of the tower with \(p=0.5\) by building from the top. That is, use tables to estimate the limit of the sequence defined recursively by (1) with \(p=0.5 .\) Estimate the maximum value of \(p > 0\) for which the sequence has a limit. b. Estimate the value of the tower with \(p=1.2\) by building from the bottom. That is, use tables to estimate the limit of the sequence defined recursively by (2) with \(p=1.2 .\) Estimate the maximum value of \(p > 1\) for which the sequence has a limit.

The Fibonacci sequence \(\\{1,1,2,3,5,8,13, \ldots\\}\) is generated by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1\). a. It can be shown that the sequence of ratios of successive terms of the sequence \(\left\\{\frac{f_{n+1}}{f_{n}}\right\\}\) has a limit \(\varphi .\) Divide both sides of the recurrence relation by \(f_{n},\) take the limit as \(n \rightarrow \infty,\) and show that \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}}=\frac{1+\sqrt{5}}{2} \approx 1.618\). b. Show that \(\lim _{n \rightarrow \infty} \frac{f_{n-1}}{f_{n+1}}=1-\frac{1}{\varphi} \approx 0.382\). c. Now consider the harmonic series and group terms as follows: $$\sum_{k=1}^{\infty} \frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\left(\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)$$ $$+\left(\frac{1}{9}+\cdots+\frac{1}{13}\right)+\cdots$$ With the Fibonacci sequence in mind, show that $$\sum_{k=1}^{\infty} \frac{1}{k} \geq 1+\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{3}{8}+\frac{5}{13}+\cdots=1+\sum_{k=1}^{\infty} \frac{f_{k-1}}{f_{k+1}}.$$ d. Use part (b) to conclude that the harmonic series diverges. (Source: The College Mathematics Journal, 43, May 2012)

An insulated window consists of two parallel panes of glass with a small spacing between them. Suppose that each pane reflects a fraction \(p\) of the incoming light and transmits the remaining light. Considering all reflections of light between the panes, what fraction of the incoming light is ultimately transmitted by the window? Assume the amount of incoming light is 1.

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