/*! 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 60 Which series converge, and which... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 60

Which series converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$\sum_{n=1}^{\infty}\left(1-\frac{1}{n}\right)^{n}$$

Short Answer

Expert verified
The series diverges because \(\lim_{n \to \infty} a_n = \frac{1}{e} \neq 0\).

Step by step solution

01

Understand the General Term

The general term of the series is \(a_n = \left(1 - \frac{1}{n}\right)^n\). As \(n\) increases, the term inside the parentheses approaches 1, but the exponent also increases. We need to analyze \(a_n\) to determine the series behavior.
02

Examine the Limit of the General Term

Consider \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n \). By using the fact that \( \lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n = \frac{1}{e} \) as derived from the exponential function definition, we see that each \(a_n\) approaches \(\frac{1}{e}\), which is not zero.
03

Apply the Basic Divergence Test

The basic divergence test (or the \(n\)-th term test for divergence) states that if \( \lim_{n \to \infty} a_n eq 0 \), the series \(\sum a_n\) diverges. Since we found that \(\lim_{n \to \infty} a_n = \frac{1}{e} eq 0\), the series \(\sum_{n=1}^{\infty} \left(1 - \frac{1}{n}\right)^n\) diverges.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

General Term
In a series, the term "general term" refers to the expression that denotes all the terms in the sequence that are being summed. For our exercise, the general term is given by \( a_n = \left(1 - \frac{1}{n}\right)^n \). This expression will produce each of the individual terms in the series as \( n \) changes. By evaluating this term as \( n \) increases, we can gain insight into the behavior of the series.
  • Start by substituting small integers into the general term formula to see the pattern.
  • The expression inside the parenthesis, \( 1 - \frac{1}{n} \), approaches 1 as \( n \) becomes very large.
  • The entire expression is affected by the exponent \( n \), influencing the rate of approach to a certain value.
Analyzing the general term is crucial because it sets the stage for determining whether the series converges or diverges.
Limit of a Series
To determine how a series behaves as \( n \) tends to infinity, we consider the limit of its general term. In the given problem, we must evaluate \[ \lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n. \] This limit results from a well-known limit in calculus that is crucial for understanding exponential growth.
  • If the limit of the general term equals zero, the series may converge.
  • If the limit is not zero, the series will definitely diverge due to the divergence test.
For our series, this limit equals \( \frac{1}{e} \), which is approximately 0.3679. It's important to note that since this is not zero, we move on to the divergence test to finalize the conclusion.
Divergence Test
The divergence test is a fundamental tool used to determine if an infinite series diverges. The test, also known as the \( n \)-th term test for divergence, states:
  • If the limit of the general term \( \lim_{n \to \infty} a_n eq 0 \), then the series diverges.
  • If \( \lim_{n \to \infty} a_n = 0 \), the series might converge, but further testing is needed.
In our exercise, we concluded that \[ \lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n = \frac{1}{e}, \] which is clearly not zero. Therefore, applying the divergence test confirms that the series diverges, meaning it does not have a finite sum. This step is crucial for quickly identifying series that cannot converge.
Exponential Function
The exponential function is a powerful concept in calculus, and it plays a significant role in understanding the behavior of growth and decay in mathematical series. The expression \( \left(1 - \frac{1}{n}\right)^n \) as \( n \to \infty \) is related to the exponential function \( e^{-1} \).
  • The form \( \left(1 + \frac{x}{n}\right)^n \) approaches \( e^x \) as \( n \to \infty \).
  • For our series \( \left(1 - \frac{1}{n}\right)^n \), this is equivalent to \( e^{-1} = \frac{1}{e} \).
Understanding this behavior is essential because it allows us to compute limits accurately, that are pivotal in determining the convergence or divergence of a series.Using the properties of exponential functions equips us with the tools to tackle intricate behavior patterns in sequences and series effortlessly.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that the error \(\left(L-s_{n}\right)\) obtained by replacing a convergent geometric series with one of its partial sums \(s_{n}\) is \(a r^{n} /(1-r)\)

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$a_{n}=\frac{\ln n}{n}$$

Taylor's formula with \(n=1\) and \(a=0\) gives the linearization of a function at \(x=0 .\) With \(n=2\) and \(n=3\) we obtain the standard quadratic and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions: a. For what values of \(x\) can the function be replaced by each approximation with an error less than \(10^{-2} ?\). b. What is the maximum error we could expect if we replace the function by each approximation over the specified interval? Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals. Step \(I:\) Plot the function over the specified interval. Step 2: Find the Taylor polynomials \(P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) at \(x=0\) Step 3: Calculate the \((n+1)\) st derivative \(f^{(n+1)}(c)\) associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of \(c\) over the specified interval and estimate its maximum absolute value, \(M .\) Step 4: Calculate the remainder \(R_{n}(x)\) for each polynomial. Using the estimate \(M\) from Step 3 in place of \(f^{(n+1)}(c),\) plot \(R_{n}(x)\) over the specified interval. Then estimate the values of \(x\) that answer question (a). Step 5: Compare your estimated error with the actual error \(E_{n}(x)=\left|f(x)-P_{n}(x)\right|\) by plotting \(E_{n}(x)\) over the specified interval. This will help answer question (b). Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. $$f(x)=e^{-x} \cos 2 x, \quad|x| \leq 1$$

Find the first four nonzero terms in the Maclaurin series for the functions. $$\sin \left(\tan ^{-1} x\right)$$

In each of the geometric series, write out the first few terms of the series to find \(a\) and \(r\), and find the sum of the series. Then express the inequality \(|r|<1\) in terms of \(x\) and find the values of \(x\) for which the inequality holds and the series converges. $$\sum_{n=0}^{\infty} 3\left(\frac{x-1}{2}\right)^{n}$$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.