/*! 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 4 Determine whether the series con... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the series converges or diverges. For convergent series, find the sum of the series. $$\sum_{i=0}^{\infty} 4\left(\frac{1}{2}\right)^{k}$$

Short Answer

Expert verified
The series \( \sum_{i=0}^{\infty} 4\left(\frac{1}{2}\right)^{k} \) converges and the sum of the series is 8.

Step by step solution

01

Identify the Common Ratio

First, note the common ratio \(r\) in the series. Looking at the original series, it's clear to see that the common ratio is \( \frac{1}{2} \).
02

Determine Convergence or Divergence

A geometric series will converge if the absolute value of the common ratio is less than 1, as the terms in the series will get smaller and smaller, eventually approaching 0. Since \( | \frac{1}{2} | < 1 \), we can say that the series \( \sum_{i=0}^{\infty} 4\left(\frac{1}{2}\right)^{k} \) converges.
03

Compute the Sum

Finally, we can use the formula for the sum \( S \) of an infinite geometric series: \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. Using this formula the sum of the series is: \( S = \frac{4}{1 - \frac{1}{2}} = 8 \).

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Key Concepts

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

Series Convergence
Understanding when a series converges is critical in mathematics, especially when dealing with infinite series. Convergence refers to the behavior of a series as the number of terms grows without bound. Specifically, if the series approaches a specific value as the number of terms increases, it is said to converge. This is different from divergence, where the terms don't settle towards any value but instead continue to grow or decrease without approaching a specific point.

In the case of the given series, we determine convergence by looking at the behavior of the terms as we add more of them. If each additional term adds an increasingly smaller amount to the sum, such that we can predict the series is getting closer to a finite number, then we say the series converges. The series provided in the exercise, an infinite geometric series, is a perfect candidate for this analysis.
Common Ratio
The common ratio, denoted as \( r \), plays a pivotal role in the behavior of geometric series. It is the factor by which consecutive terms of the series are multiplied. To find the common ratio, simply take any term in the series (after the first) and divide it by the preceding term. Consistency of the ratio across all terms is a defining characteristic of geometric series.

For example, in our series \( 4\frac{1}{2}^{k} \), the common ratio is \( \frac{1}{2} \). The value of this common ratio determines whether the series converges or diverges. A common ratio with an absolute value less than 1 indicates a converging geometric series, while a value greater than 1 indicates divergence. This is because, with each successive term, the impact on the total sum diminishes when \| r \| < 1, leading to a fixed limit.
Infinite Series
An infinite series is a sum of infinite terms that follow a certain pattern or rule. Not every infinite series has a finite sum, but when they do, they provide remarkable insights into the behavior of seemingly endless processes. The question of whether an infinite series converges to a finite value is foundational in the study of sequences and series.

Geometric series are a type of infinite series where each term after the first is found by multiplying the previous term by a constant (the common ratio). Our exercise involves an infinite geometric series, where terms continue indefinitely. Whether such a series can be summed depends on its convergence. As shown in the solution, with a common ratio of \( \frac{1}{2} \), the terms get closer to zero, implying that a finite sum is indeed possible.
Geometric Series Sum
When a geometric series converges, it is possible to find its sum using a straightforward formula. The sum of an infinite geometric series is given by \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. This powerful formula comes from the idea that the difference between the series and itself shifted by one term converges to the first term, allowing us to solve for the sum \( S \).

In the given exercise, the first term of the series is 4 and the common ratio is \( \frac{1}{2} \). By applying the sum formula, we can calculate that the series converges to a sum of 8. It's important to note that this formula only applies when \| r \| < 1, as only then does the series converge to a finite limit. The simplicity yet effectiveness of this formula is a beautiful aspect of geometric series.

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Most popular questions from this chapter

Suppose that \(a_{1}=1\) and \(a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{4}{a_{n}}\right) .\) Show numerically that the sequence converges to \(2 .\) To find this limit analytically, $$\text { let } L=\lim _{n \rightarrow \infty} a_{n+1}=\lim _{n \rightarrow \infty} a_{n}$$ and solve the equation $$L=\frac{1}{2}\left(L+\frac{4}{L}\right)$$

The cutoff frequency setting on a music synthesizer has a dramatic effect on the timbre of the tone produced. In terms of harmonic content (see exercise 42 ), when the cutoff frequency is set at \(n>0,\) all harmonics beyond the \(n\) th harmonic are set equal to \(0 .\) In Fourier series terms, explain how this corresponds to the partial sum \(F_{n}(x)\). For the sawtooth and square waves, graph the waveforms with the cutoff frequency set at 4 Compare these to the waveforms with the cutoff frequency set at \(2 .\) As the setting is lowered, you hear more of a "pure" tone. Briefly explain why.

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 \(\operatorname{sum} \frac{10}{9}\)

We have seen that \(\sin 1=1-\frac{1}{31}+\frac{1}{51}+\cdots .\) Determine how many terms are needed to approximate sin 1 to within \(10^{-5}\) Show that \(\sin 1=\int_{0}^{1} \cos x d x .\) Determine how many points are needed for Simpson's Rule to approximate this integral to within \(10^{-5} .\) Compare the efficiency of Maclaurin series and Simpson's Rule for this problem.

Find the Taylor series for \(e^{x}\) about a general center \(c\)

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