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Choose your test Use the test of your choice to determine whether the following series converge. $$\sum_{k=2}^{\infty} 100 k^{-k}$$

Short Answer

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Explain your reasoning. Answer: The series $$\sum_{k=2}^{\infty} 100 k^{-k}$$ converges. We used the Ratio Test, which involves computing the limit of the absolute ratio of consecutive terms, i.e., $$\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right|$$, where \(a_k = 100 k^{-k}\). After computing the limit, we obtained $$\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| = \frac{1}{e}$$ Since this value is less than 1, the given series converges according to the Ratio Test.

Step by step solution

01

Write down the terms of the series and the Ratio Test formula

The terms of the series are given by \(a_k = 100 k^{-k}\) and the formula for the Ratio Test is: $$\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right|$$
02

Find the ratio of consecutive terms

First, find the expression for \(a_{k+1}/a_k\): $$\frac{a_{k+1}}{a_k} = \frac{100(k+1)^{-(k+1)}}{100 k^{-k}} = \frac{(k+1)^{-(k+1)}}{k^{-k}}$$
03

Simplify the expression

Rewrite the expression as: $$\frac{(k+1)^{-(k+1)}}{k^{-k}} = \left(\frac{k}{k+1}\right)^k$$
04

Compute the limit

Next, compute the limit as \(k\) approaches infinity: $$\lim_{k\to\infty} \left(\frac{k}{k+1}\right)^k$$ To compute this limit, we can make a substitution: let \(n=k+1\), then \(k=n-1\) and \(k\to\infty\) as \(n\to\infty\). So, the limit becomes: $$\lim_{n\to\infty} \left(\frac{n-1}{n}\right)^{n-1}$$ Recognize that this limit is the exponent only in the limit definition of Euler's number \(e\) and is known to be: $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e$$ And we have our current limit: $$\lim_{n\to\infty} \left(\frac{n-1}{n}\right)^{n-1}=\frac{1}{e}$$
05

Conclude the result

Since the limit of the absolute ratio of consecutive terms is less than 1 (\(\frac{1}{e}<1\)), we can conclude that the given series converges. $$\sum_{k=2}^{\infty} 100 k^{-k} \text{ converges}$$

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Key Concepts

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

Ratio Test
The Ratio Test is a valuable technique used to determine whether a series converges or diverges. This test is particularly useful for series whose terms involve factorials, exponentials, or powers, making it generally applicable for complex series. To apply the Ratio Test, take the following steps:
  • Identify the series and its terms, often denoted as \(a_k\).
  • Compute the ratio of successive terms, \(\frac{a_{k+1}}{a_k}\).
  • Take the limit of the absolute value of this ratio as \(k\) approaches infinity: \(\lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right|\).
If this limit is less than 1, the series converges. If it is greater than 1, the series diverges. If the limit equals 1, the test is inconclusive, meaning we need another method to determine convergence.
Series
A series is essentially the sum of the terms of a sequence. It may continue indefinitely, forming what is called an infinite series. Understanding series is crucial in calculus and analysis, where they model various phenomena.In mathematical notation, a series is often written using the sigma notation \(\sum_{k=1}^{\infty} a_k\), indicating that terms of the sequence \(a_k\) are added up from \(k=1\) to infinity. Exploring the behavior of series involves:
  • Examining convergence, determining if the sum approaches a finite value as more terms are added.
  • Exploring divergence, where the sum increases without bound or oscillates.
  • Applying tests such as the Ratio Test, Root Test, or Integral Test to understand the behavior of the series.
The study of series is essential for advanced calculus topics such as Taylor and Maclaurin series, which approximate functions.
Exponential Growth
Exponential growth describes a process where quantities increase at a rate proportional to their current value. This type of growth is common in various fields, such as biology, finance, and technology.The exponential function \(e^x\) is central to understanding exponential growth because it models continuous compounding. For instance, if a population doubles every fixed time period, it's growing exponentially. Key characteristics include:
  • Constant percentage growth per time unit.
  • Use of mathematical models to predict future sizes.
  • Limitless theoretical expansion under ideal conditions.
In mathematics, analyzing series involving exponential terms can reveal if a process exhibits growth or decay patterns, which can be crucial for applications like interest calculations or natural phenomena modeling.
Euler's Number
Euler's number, denoted as \(e\), is approximately 2.71828 and is a fundamental constant in mathematics. This irrational number arises naturally in many areas of mathematics, particularly in calculus, through the study of limits and exponential functions.An insightful property of \(e\) is its unique relationship with exponential functions, often expressed as \(e^x\). Key points about \(e\) include:
  • The limit \(\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e\) demonstrates its properties in continuous compounding and growth rates.
  • \(e\) forms the basis of natural logarithms, connecting multiplicative rates to linear processes.
  • Euler's number appears in the derivatives and integrals of exponential functions, making it essential for solving problems involving growth, decay, and oscillations.
Understanding \(e\) and its applications opens numerous mathematical doors, allowing easier interpretations and solutions of complex problems, particularly those involving limits and change rates.

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Most popular questions from this chapter

A glimpse ahead to power series Use the Ratio Test to determine the values of \(x \geq 0\) for which each series converges. $$\sum_{k=1}^{\infty} \frac{x^{k}}{k}$$

Find a series that a. converges faster than \(\sum \frac{1}{k^{2}}\) but slower than \(\sum \frac{1}{k^{3}}\) b. diverges faster than \(\sum \frac{1}{k}\) but slower than \(\sum \frac{1}{\sqrt{k}}\) c. converges faster than \(\sum \frac{1}{k \ln ^{2} k}\) but slower than \(\sum \frac{1}{k^{2}}\)

Explain the fallacy in the following argument. Let \(x=\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\) and \(y=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots .\) It follows that \(2 y=x+y\), which implies that \(x=y .\) On the other hand, $$x-y=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\cdots>0$$ is a sum of positive terms, so \(x>y .\) Thus, we have shown that \(x=y\) and \(x>y\).

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty} 9(0.1)^{k}$$

Suppose an alternating series \(\sum(-1)^{k} a_{k}\) converges to \(S\) and the sum of the first \(n\) terms of the series is \(S_{n}\) Suppose also that the difference between the magnitudes of consecutive terms decreases with \(k\). It can be shown that for \(n \geq 1,\) $$\left|S-\left[S_{n}+\frac{(-1)^{n+1} a_{n+1}}{2}\right]\right| \leq \frac{1}{2}\left|a_{n+1}-a_{n+2}\right|$$ a. Interpret this inequality and explain why it gives a better approximation to \(S\) than simply using \(S_{n}\) to approximate \(S\). b. For the following series, determine how many terms of the series are needed to approximate its exact value with an error less than \(10^{-6}\) using both \(S_{n}\) and the method explained in part (a). (i) \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\) (ii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln k}\) (iii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)

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