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Prove that for \(a_{k}>0, \sum_{k=1}^{\infty} a_{k}\) converges if and only if \(\sum_{k=1}^{\infty} \frac{a_{k}}{1+a_{k}}\) converges. (Hint: If \(x<1\), then \(x<\frac{2 x}{1+x}\).)

Short Answer

Expert verified
The presented series \(\sum_{k=1}^{\infty} a_{k}\) and \(\sum_{k=1}^{\infty} \frac{a_{k}}{1+a_{k}}\) are shown to be both conditions of each other respectively, that is, each series converges if and only if the other does as well. This was established through applying comparison test and utilizing given hint.

Step by step solution

01

Formulate the If Direction

We first assume that \(\sum_{k=1}^{\infty} a_{k}\) converges. We need to prove that under this assumption, \(\sum_{k=1}^{\infty} \frac{a_{k}}{1+a_{k}}\) also converges.
02

Apply the Comparison Test in the If Direction

In order to apply comparison test, it must be determined whether \(\frac{a_{k}}{1+a_{k}}\) is less than or equivalent to \(a_{k}\) for all \(k\). From the hint, we know that \(x<\frac{2 x}{1+x}\) for \(x<1\). By replacing \(2x\) with \(a_{k}\), we have \(\frac{a_{k}}{1+a_{k}}<a_{k}\). Thus, if \(\sum_{k=1}^{\infty} a_{k}\) converges, then \(\sum_{k=1}^{\infty} \frac{a_{k}}{1+a_{k}}\) must also converge by comparison test.
03

Formulate the Only If Direction

Now assume that \(\sum_{k=1}^{\infty} \frac{a_{k}}{1+a_{k}}\) converges. We need to prove that under this assumption, \(\sum_{k=1}^{\infty} a_{k}\) also converges.
04

Apply the Comparison Test in the Only If Direction

In order to apply comparison test, it must be determined whether \(a_{k}\) is less than or equivalent to some constant times \(\frac{a_{k}}{1+a_{k}}\) for all \(k\). Applying the hint again, we see that \(a_{k}=\frac{a_{k}}{1}\leq\frac{2 a_{k}}{1+a_{k}}\). Therefore, by the comparison test, if \(\sum_{k=1}^{\infty} \frac{a_{k}}{1+a_{k}}\) converges, then \(\sum_{k=1}^{\infty} a_{k}\) must also converge.

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Key Concepts

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

Comparison Test
Understanding the comparison test is essential for analyzing the convergence of series in calculus. It's a method to determine whether an infinite series converges or diverges by comparing it to another series whose convergence properties are known. The test is based on the idea that if the absolute value of the terms of series A is always less than or equal to the terms of series B for all n, and if series B converges, then series A also converges. Conversely, if series A diverges, then any series larger than A also diverges.

Let's use the comparison test effectively: given two series \(\sum_{n=1}^{\infty} a_n\) and \(\sum_{n=1}^{\infty} b_n\), if \(0 \leq a_n \leq b_n\) for all n and \(\sum_{n=1}^{\infty} b_n\) is a convergent series, then \(\sum_{n=1}^{\infty} a_n\) is also convergent. This concept is instrumental when analyzing series that are not immediately apparent whether they converge or diverge.
Infinite Series
In calculus, an infinite series is a sum of infinite terms that successively follow a specific rule. It is written as \(\sum_{n=1}^{\infty} a_n\), where \(a_n\) are the terms of the series. Infinite series are significant because they can represent a variety of mathematical and physical concepts, such as the value of \(\pi\), the exponential function, and many more. Understanding infinite series allows for the analysis of complex problems in several branches of mathematics and physics.

However, not all infinite series converge to a finite value; some grow without bound or do not settle to a particular value. Determining convergence or divergence is a key challenge in working with such series, which involves various tests like the comparison test, ratio test, and root test, among others.
Convergent Series
A series is considered convergent if the sum of its infinite terms approaches a specific finite value as the number of terms grows without bound. This concept is a cornerstone of calculus and analysis because it provides a way to deal with infinite processes by examining their end behavior. For example, when we say the series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) converges, we mean that as we take the sum of more and more terms, it approaches a particular number, which in this case is \(\frac{\pi^2}{6}\).

Convergence can sometimes be determined through direct computation, but often we need to use tests like the comparison test explained earlier. A deep understanding of when and why series converge is crucial for their practical application, such as in Fourier series, which are used for signal processing and in the solving of differential equations.
Calculus
Calculus is a branch of mathematics that deals with the study of change and motion. Its two main subfields are differential calculus, concerning rates of change and slopes of curves, and integral calculus, which concerns accumulation of quantities and the areas under and between curves. Infinite series come into play in integral calculus when we talk about the representation of functions as power series and in the evaluation of integrals that cannot be expressed in terms of elementary functions.

In terms of series, calculus provides us with the tools to analyze their convergence, as seen in our earlier example regarding the comparison test. These methods are not only theoretical but also have practical applications in fields like physics, engineering, economics, and even in complex algorithms used in computer science.

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Most popular questions from this chapter

Use the Binomial Theorem to approximate the value to within \(10^{-6}\) (a) \(\frac{2}{\sqrt[3]{9}}\) (b) \(\sqrt[4]{17}\)

If \(n\) and \(k\) are positive integers with \(n>k,\) show that \(\left(\begin{array}{l}n \\ k\end{array}\right)=\frac{n !}{k !(n-k) !}\)

This problem is sometimes called the coupon collectors" problem. The problem is faced by collectors of trading cards. If there are \(n\) different cards that make a complete set and you randomly obtain one at a time, how many cards would you expect to obtain before having a complete set? (By random, we mean that each different card has the same probability of \(\frac{1}{n}\) of being the next card obtained.) In exercises \(69-72,\) we find the answer for \(n=10 .\) The first step is simple; to collect one card you need to obtain one card. Now, given that you have one card, how many cards do you need to obtain to get a second (different) card? If you're lucky, the next card is it (this has probability 10). But your next card might be a duplicate, then you get a new card (this has probability \(\left.\frac{1}{10} \cdot \frac{9}{10}\right) .\) Or you might get two duplicates and then a new card (this has probability \(\left.\frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10}\right) ;\) and so on. The mean is \(1 \cdot \frac{9}{10}+2 \cdot \frac{1}{10} \cdot \frac{9}{10}+3 \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10}+\cdots\) or \(\sum_{k=1}^{\infty} k\left(\frac{1}{10}\right)^{k-1}\left(\frac{9}{10}\right)=\sum_{k=1}^{\infty} \frac{9 k}{10^{k}} .\) Using the same trick as in exercise \(68,\) show that this is a convergent series with \(\operatorname{sum} \frac{10}{9}\)

Define $$a_{n}=\sum_{k=1}^{n} \frac{1}{n+k}$$. By thinking of \(a_{n}\) as a Riemann sum, identify the definite integral to which the sequence converges.

Find the first five terms in the Taylor series about \(c=0\) for \(f(x)=e^{x} \sin x\) and compare to the product of the Taylor polynomials about \(c=0\) of \(e^{x}\) and \(\sin x\)

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