/*! 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 56 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. $$ a_{n}=\frac{(-3)^{n}}{n !} $$

Short Answer

Expert verified
The sequence converges to 0.

Step by step solution

01

Understand the Sequence

The given sequence is \( a_n = \frac{(-3)^n}{n!} \). This sequence includes an exponential term \((-3)^n\) and a factorial term \(n!\). To determine convergence or divergence, we should consider the behavior of both factors.
02

Consider the Factorial Growth

The factorial \(n!\) grows very rapidly as \(n\) increases. This factor tends to infinity much faster than exponential terms increase.
03

Consider the Exponential Term

The exponential term \((-3)^n\) has an absolute value \((3)^n\), which grows exponentially. Nevertheless, factorial growth outpaces exponential growth quickly for large \(n\).
04

Apply the Ratio Test

To better assess convergence, use the Ratio Test. Find \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(-3)^{n+1}}{(n+1)!} \cdot \frac{n!}{(-3)^n} \right| = \lim_{n \to \infty} \frac{3}{n+1} . \]
05

Evaluate the Limit

Evaluate the limit \( L = \lim_{n \to \infty} \frac{3}{n+1} = 0 \). Since \( L < 1 \), the Ratio Test confirms the sequence converges.
06

Determine the Limit

With the sequence confirmed as convergent, examine the sequence in the limit as \(n\) approaches infinity. Since the exponential term becomes negligible compared to the factorial, \[ \lim_{n \to \infty} a_n = 0. \]

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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 useful tool for determining whether a sequence or series converges. Its primary function is to compare the size of consecutive terms in a sequence as one approaches infinity. Applying the Ratio Test involves calculating a specific limit, known as L. The formula for this is: \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \] Here, \(a_n\) represents the terms in the sequence you're examining. If the ratio \( L < 1 \), the sequence converges. If \( L > 1 \), it diverges. In cases where \( L = 1 \), the test is inconclusive.
  • For the sequence \( a_n = \frac{(-3)^n}{n!} \), the Ratio Test shows \( L = \lim_{n \to \infty} \frac{3}{n+1} = 0 \), which indicates convergence.
This test effectively demonstrates how growth rates impact limits, leveraging the factorial's rapid acceleration compared to powers of negative or positive bases.
Exponential Growth
Exponential growth describes a situation where something increases at a rate proportional to its current value, signifying multiplicative growth. In mathematical sequences, this is represented by terms like \((-3)^n\). Even though exponential growth can be rapid, it's important to compare it against other growth measures in sequences.

In the sequence \( a_n = \frac{(-3)^n}{n!} \), the term \((-3)^n\) shows exponential growth. However, when coupled with a factorial in the denominator, it poses an interesting observation as factorial growth far surpasses exponential growth for larger \(n\).
  • This comparison underlines why the sequence ultimately converges: the suppressive power of the factorial overpowers the exponential component as \(n\) becomes very large.
Extended understanding of such contrasts helps students appreciate how different mathematical growth factors interact.
Factorial
A factorial, denoted as \( n! \), is the product of all positive integers up to \( n \). Essentially, \( n! = n \times (n-1) \times \dots \times 1 \). Factorials grow incredibly fast, much faster than simple powers. Due to their rapid expansion, they are often dominating elements in sequences or series where terms are both multiplying and dividing.

In the sequence \( a_n = \frac{(-3)^n}{n!} \), the \( n! \) in the denominator grows faster than the numerator \((-3)^n\).
  • This quick growth is what causes the terms of the sequence to diminish quickly as \( n \) increases, ensuring that the sequence converges to zero.
Understanding factorials is crucial in comprehending why many sequences involving factorials tend to converge.
Sequence Limit
The concept of a sequence limit is pivotal in understanding whether a sequence converges. A sequence converges to a limit if, as \( n \) increases indefinitely, the terms in the sequence approach and get arbitrarily close to a specific number. For the sequence \( a_n = \frac{(-3)^n}{n!} \), identifying the limit involves observing the terms' behavior as \( n \) approaches infinity.

The Ratio Test showed the sequence converges, and evaluation further suggests the limit is zero since \( n! \) grows faster than \( (-3)^n \).
  • Thus, \( \lim_{n\to\infty} a_n = 0 \), meaning the terms get ever closer to zero as \( n \) increases.
Appreciating how term-by-term analysis and convergence tests pinpoint sequence limits is fundamental in mathematical analysis and calculus.

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Most popular questions from this chapter

Find the Maclaurin series of \(f\) (by any method) and its radius of convergence. Graph \(f\) and its first few Taylor polynomials on the same screen. What do you notice about the relationship between these polynomials and \(f ?\) $$ f(x)=\cos \left(x^{2}\right) $$

Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number \(a\). (b) Use Taylor's Inequality to estimate the accuracy of the approximation \(f(x)=T_{x}(x)\) when \(x\) lies in the given interval. (c) Check your result in part (b) by graphing \(\left|R_{n}(x)\right|\). $$ f(x)=\ln (1+2 x), \quad a=1, \quad n=3, \quad 0.5 \leqslant x \leqslant 1.5 $$

If a surveyor measures differences in elevation when making plans for a highway across a desert, corrections must be made for the curvature of the earth. (a) If \(R\) is the radius of the earth and \(L\) is the length of the highway, show that the correction is \(C=R \sec (L / R)-R\) (b) Use a Taylor polynomial to show that \(C \approx \frac{L^{2}}{2 R}+\frac{5 L^{4}}{24 R^{3}}\) (c) Compare the corrections given by the formulas in parts (a) and (b) for a highway that is \(100 \mathrm{km}\) long. (Take the radius of the earth to be \(6370 \mathrm{km}\).)

Use the power series for \(\tan ^{-1} x\) to prove the following expression for \(\pi\) as the sum of an infinite series: $$ \pi=2 \sqrt{3} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) 3^{n}} $$

The resistivity \(\rho\) of a conducting wire is the reciprocal of the conductivity and is measured in units of ohm-meters \((\Omega-\mathrm{m}) .\) The resistivity of a given metal depends on the temperature according to the equation \(\rho(t)=\rho_{20} e^{\alpha(t-2 \omega)}\) where \(t\) is the temperature in \(^{\circ} \mathrm{C}\). There are tables that list the values of \(\alpha\) (called the temperature coefficient) and \(\rho_{20}\) (the resistivity at \(20^{\circ} \mathrm{C}\) ) for various metals. Except at very low temperatures, the resistivity varies almost linearly with temperature and so it is common to approximate the expression for \(\rho(t)\) by its first-or second-degree Taylor polynomial at \(t=20 .\) $$ \begin{array}{l}{\text { (a) Find expressions for these linear and quadratic }} \\ {\text { approximations. }} \\ {\text { (b) For copper, the tables give } \alpha=0.0039 /^{\circ} \mathrm{C} \text { and }} \\ {\rho_{20}=1.7 \times 10^{-8} \Omega \text { -m. Graph the resistivity of copper }} \\\ {\text { and the linear and quadratic approximations for }} \\ {-250^{\circ} \mathrm{C} \leqslant t \leqslant 1000^{\circ} \mathrm{C} \text { . }} \\\ {\text { (c) For what values of } t \text { does the linear approximation agree }} \\ {\text { with the exponential expression to within one percent? }}\end{array} $$

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