/*! 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 128 Consider the sequence \(\left\\{... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the sequence \(\left\\{a_{n}\right\\}=\left\\{n r^{n}\right\\}\). Decide whether \(\left\\{a_{n}\right\\}\) converges for each value of \(r\). (a) \(r=\frac{1}{2}\) (b) \(r=1\) (c) \(r=\frac{3}{2}\) (d) For what values or \(r\) does the sequence \(\left\\{n r^{n}\right\\}\) converge?

Short Answer

Expert verified
The sequence \(\left\{a_{n}\right\}=\left\{n r^{n}\right\}\) converges when \(r=\frac{1}{2}\), diverges when \(r=1\) or \(r=\frac{3}{2}\), and generally converges for all \(r<1\).

Step by step solution

01

Checking for convergence when \(r=\frac{1}{2}\)

Firstly, substitute \(r=\frac{1}{2}\) into the sequence \(\left\{a_{n}\right\}\) to get \(\left\{a_{n}\right\}=\left\{n \left(\frac{1}{2}\right)^{n}\right\}\). This sequence clearly diminishes towards zero as \(n\) increases since \(n\) is eventually overpowered by the exponential decay of \(\left(\frac{1}{2}\right)^{n}\). Therefore, \(\left\{a_{n}\right\}\) converges when \(r=\frac{1}{2}\).
02

Checking for convergence when \(r=1\)

Secondly, let's substitute \(r=1\) into the sequence \(\left\{a_{n}\right\}\) gives us \(\left\{a_{n}\right\}=\left\{n\right\}\), which clearly diverges as \(n\) keeps increasing indefinitely.
03

Checking for convergence when \(r=\frac{3}{2}\)

Next, substitute \(r=\frac{3}{2}\) into the sequence \(\left\{a_{n}\right\}\) yields \(\left\{a_{n}\right\}=\left\{n \left(\frac{3}{2}\right)^{n}\right\}\). Since the \(n\) in the numerator grows slower than the exponential term in the denominator, \(\left\{a_{n}\right\}\) diverges as \(n\) increases.
04

Determining the convergence conditions for general \(r\)

From the above results, it can be deduced that the sequence \(\left\{a_{n}\right\}=\left\{n r^{n}\right\}\) converges if \(r<1\) and diverges if \(r\geq 1\). Therefore, the sequence \(\left\{a_{n}\right\}\) converges if, and only if, \(r<1\).

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Key Concepts

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

Sequence Convergence Calculus
Sequence convergence is a fundamental concept in calculus, particularly in the study of infinite sequences. It deals with understanding whether a sequence of numbers tends to a specific limit as the number of terms goes to infinity. In mathematical terms, a sequence \( \{a_n\} \) is said to converge to a real number \( L \) if, for every positive number \( ε \) (no matter how small), there is a corresponding integer \( N \) such that for all \( n \geq N \) the terms of the sequence satisfy \( |a_n - L| < ε \).

To put it simply, after some point in the sequence, all further terms get arbitrarily close to the limit \( L \) and stay close. Checking for convergence involves looking at the terms of a sequence and determining if they stabilize or spread out as the sequence progresses.

In our exercise scenario, we consider the sequence \( \{n r^n\} \) for different values of \( r \). A key aspect in determining the convergence of such a sequence is to analyze the behavior of \( r^n \) as \( n \) tends to infinity. If \( r < 1 \) the terms \( r^n \) get smaller as \( n \) increases, implying potential convergence. Conversely, if \( r \geq 1 \) the terms either stabilize or increase, and the sequence is less likely to converge. Thus, calculus provides the tools to make these precise assessments and guides us towards understanding the limiting behavior of sequences.
Geometric Series Convergence
The concept of geometric series convergence is closely related to sequence convergence. A geometric series is the sum of the terms of a geometric sequence, where each term after the first is found by multiplying the previous term by a constant called the ratio \( r \). The geometric series is expressed as \( S = a_1 + a_1r + a_1r^2 + \.\.\. \) where \( a_1 \) is the first term. Convergence of this series occurs if the absolute value of the ratio is less than one, that is, \( |r| < 1 \).

When a geometric series converges, its sum approaches a finite limit, specifically \( S = \frac{a_1}{1 - r} \) for \( |r| < 1 \). If \( |r| \geq 1 \) the series diverges, which means that the sum goes off to infinity and does not settle at any particular value.

In the context of our exercise, when analyzing the convergence of the sequence \( \{n r^n\} \), it's very much like looking at the behaviors of a geometric series with an extra factor of \( n \) thrown in the mix. The convergence traits of geometric sequences set the underlying pattern while this additional \( n \) factor influences the terms' growth rate. Understanding this underlying geometric nature can be crucial in evaluating the overall convergence behavior of similar sequences, as the ratio \( r \) dictates if the terms are drawing closer together or growing apart unbounded.
Exponential Decay
Exponential decay is a concept often used to describe processes that decrease at a rate proportional to their current value. In mathematics, this translates to functions of the form \( f(t) = Ce^{-kt} \) where \( C \) is a constant, \( t \) is a variable (often representing time), and \( k \) is a positive constant representing the rate of decay. This concept is also present in sequences where terms decrease exponentially, typically at a rate determined by a base \( r \) where \( 0 < r < 1 \).

Applying this to our exercise for the case where \( r = \frac{1}{2} \) in the sequence \( \{n r^n\} \), the term \( (\frac{1}{2})^n \) illustrates an exponential decay because the base of the exponent is a fraction between 0 and 1. As \( n \) grows, the value \( (\frac{1}{2})^n \) becomes smaller and smaller, decreasing more rapidly than the linear increase of \( n \). This guarantees the overall terms of the sequence will diminish to zero, hence showing convergence.

Understanding exponential decay is vital in many fields, from physics to finance, and recognizing it in mathematical sequences is a critical skill. By identifying the exponential decay in a sequence, we can predict the behavior of the sequence over time and determine whether or not it will converge to a finite limit or decay to zero.

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Most popular questions from this chapter

Fibonacci Sequence In a study of the progeny of rabbits, Fibonacci (ca. \(1170-\) ca. 1240 ) encountered the sequence now bearing his name. It is defined recursively by \(a_{n+2}=a_{n}+a_{n+1}, \quad\) where \(\quad a_{1}=1\) and \(a_{2}=1\). (a) Write the first 12 terms of the sequence. (b) Write the first 10 terms of the sequence defined by $$ b_{n}=\frac{a_{n+1}}{a_{n}}, \quad n \geq 1 . $$ (c) Using the definition in part (b), show that $$ b_{n}=1+\frac{1}{b_{n-1}} $$ (d) The golden ratio \(\rho\) can be defined by \(\lim _{n \rightarrow \infty} b_{n}=\rho\). Show that \(\rho=1+1 / \rho\) and solve this equation for \(\rho\).

Find all values of \(x\) for which the series converges. For these values of \(x\), write the sum of the series as a function of \(x\). $$ \sum_{n=1}^{\infty}(x-1)^{n} $$

Consider the sequence \(\left\\{a_{n}\right\\}=\left\\{\frac{1}{n} \sum_{k=1}^{n} \frac{1}{1+(k / n)}\right\\}\). (a) Write the first five terms of \(\left\\{a_{n}\right\\}\). (b) Show that \(\lim _{n \rightarrow \infty} a_{n}=\ln 2\) by interpreting \(a_{n}\) as a Riemann sum of a definite integral.

Show that the Maclaurin series of the function \(g(x)=\frac{x}{1-x-x^{2}}\) is \(\sum_{n=1}^{\infty} F_{n} x^{n}\) where \(F_{n}\) is the \(n\) th Fibonacci number with \(F_{1}=F_{2}=1\) and \(F_{n}=F_{n-2}+F_{n-1}\), for \(n \geq 3 .\) (Hint: Write \(\frac{x}{1-x-x^{2}}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots\) and multiply each side of this equation by \(1-x-x^{2}\).)

Let \(f(n)\) be the sum of the first \(n\) terms of the sequence 0,1 , \(1,2,2,3,3,4, \ldots\), where the \(n\) th term is given by \(a_{n}=\left\\{\begin{array}{cc}n / 2, & \text { if } n \text { is even } \\\ (n-1) / 2, & \text { if } n \text { is odd }\end{array}\right.\) Show that if \(x\) and \(y\) are positive integers and \(x>y\) then \(x y=f(x+y)-f(x-y)\)

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