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

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Determine whether the sequence converges or diverges. If it converges, find the limit. $$ a_{n}=1-(0.2)^{n} $$

Short Answer

Expert verified
The sequence converges to 1.

Step by step solution

01

Identify the Sequence Components

The given sequence is \( a_n = 1 - (0.2)^n \). This sequence consists of two parts: the constant term 1, and \((0.2)^n\), which is a term that changes with \(n\).
02

Analyze the Behavior of \((0.2)^n\)

The term \((0.2)^n\) decreases as \(n\) increases because \(0.2\) is a positive number less than 1. Therefore, \((0.2)^n\) converges to 0 as \(n\) approaches infinity.
03

Determine the Limit of \(a_n\)

As \((0.2)^n\) converges to 0, we can write the sequence limit as \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} (1 - (0.2)^n) = 1 - 0 = 1 \). Therefore, the sequence converges to 1.

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Key Concepts

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

Limits of Sequences
A sequence in mathematics is simply an ordered list of numbers. One of the key ideas when studying sequences is understanding their limit. The limit of a sequence is the value that the terms of the sequence get closer to as the sequence progresses. If a sequence gets infinitely closer to a specific number, that number is called the limit of the sequence.
In the given exercise, the limit of the sequence \( a_n = 1 - (0.2)^n \) is what we are looking to determine. To find this, we observe what happens to \( (0.2)^n \) as \( n \) increases. Since \( 0.2 \) is a positive number less than 1, \( (0.2)^n \) gets smaller and smaller, approaching zero. As a result, the sequence effectively approaches a limit of \( 1 - 0 = 1 \). This concept of convergence towards a limit is central to understanding sequences.
Convergent Sequences
When we talk about convergent sequences, we refer to sequences that approach a specific value, known as the limit, as the number of terms, denoted by \( n \), grows towards infinity. Essentially, a sequence is considered convergent if there is some real number \( L \) such that the terms of the sequence become arbitrarily close to \( L \) for sufficiently large \( n \).
In our exercise, because \( (0.2)^n \) approaches zero as \( n \) increases, the sequence \( a_n = 1 - (0.2)^n \) demonstrates convergence. It's not enough for the sequence terms to just get closer to one another; they must specifically get closer to the limit value, in this case, 1. Not all sequences are convergent, but this specific sequence is an excellent example of how a function can smoothly settle into stability.
Exponential Decay
Exponential decay refers to the pattern where quantities decrease at a rate proportional to their current value, leading to an ever-faster reduction. It's a fundamental concept in sequences and can be seen clearly in sequences where the base of the exponential term is between 0 and 1.
For the sequence \( (0.2)^n \) in the exercise, as \( n \) becomes larger, the value diminishes because it follows an exponential decay pattern. Each term becomes smaller than its predecessor, approaching closer to zero in a logarithmic fashion. Because this exponential decay term is subtracted in the sequence \( a_n = 1 - (0.2)^n \), its diminishing nature is critical to why \( a_n \) converges toward 1. Understanding exponential decay helps us comprehend how and why certain sequences stabilize at a particular value over time.

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Most popular questions from this chapter

Let \(f_{n}(x)=(\sin n x) / n^{2} .\) Show that the series \(\Sigma f(x)\) converges for all values of \(x\) but the series of derivatives \(\Sigma f_{n}^{\prime}(x)\) diverges when \(x=2 n \pi, n\) an integer. For what values of \(x\) does the series \(\sum f_{n}^{\prime \prime}(x)\) converge?

The size of an undisturbed fish population has been modeled by the formula $$ p_{n+1}=\frac{b p_{n}}{a+p_{n}} $$ where \(p_{n}\) is the fish population after \(n\) years and \(a\) and \(b\) are positive constants that depend on the species and its environment. Suppose that the population in year 0 is \(p_{0} > 0\) . (a) Show that if \(\left\\{p_{n}\right\\}\) is convergent, then the only possible values for its limit are 0 and \(b-a\) . (b) Show that \(p_{n+1}<(b / a) p_{n}\) (c) Use part (b) to show that if \(a > b,\) then \(\lim _{n \rightarrow \infty} p_{n}=0\) in other words, the population dies out. (d) Now assume that \(a < b .\) Show that if \(p_{0} < b-a\) , then \(\left\\{p_{n}\right\\}\) is increasing and \(0 < p_{n}< b-a\) . Show also that if \(p_{0} >b-a\) , then \(\left\\{p_{n}\right\\}\) is decreasing and \(p_{n} > b-a\) Deduce that if \(a < b\) , then \(\lim _{n \rightarrow \infty} p_{n}=b-a\)

If \(\Sigma a_{n}\) is convergent and \(\Sigma b_{n}\) is divergent, show that the series \(\Sigma\left(a_{n}+b_{n}\right)\) is divergent. IHint: Argue by contradiction. \(]\)

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