/*! 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 27 If the sequence is convergent, f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 27

If the sequence is convergent, find its limit. If it is divergent, explain why. $$a_{n}=\frac{1}{3^{n}}$$

Short Answer

Expert verified
The sequence converges, and its limit is 0.

Step by step solution

01

Identify the Form of the Sequence

The given sequence is \( a_n = \frac{1}{3^n} \). It's a geometric sequence with the first term \(a_1 = \frac{1}{3}\) and a common ratio \(r = \frac{1}{3}\).
02

Determine the Convergence of a Geometric Sequence

A geometric sequence \(a_n = a_1 \cdot r^{n-1}\) converges if the absolute value of the common ratio \( |r| < 1 \). In this sequence, \(|r| = \left|\frac{1}{3}\right| = \frac{1}{3} < 1\), thus the sequence converges.
03

Find the Limit of the Convergent Sequence

For a geometric sequence with \(|r| < 1\), the limit as \(n\) approaches infinity is zero. Therefore, the limit of \(a_n = \frac{1}{3^n}\) is \( \lim_{{n \to \infty}} a_n = 0 \).

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.

Geometric Sequence
A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant, known as the common ratio. An example looks like this: if the first term is 2 and the common ratio is 3, the sequence would be 2, 6, 18, 54, and so on.
In the given problem, the sequence is defined by \( a_n = \frac{1}{3^n} \). This sequence is geometric because each term is obtained by multiplying the previous term by \( \frac{1}{3} \). This specific sequence starts with \( a_1 = \frac{1}{3} \), and its common ratio \( r = \frac{1}{3} \).
Things to remember about geometric sequences:
  • Every term is multiplied by the common ratio \( r \) to get the next term.
  • The formula used is \( a_n = a_1 \cdot r^{n-1} \).
  • If the common ratio is a fraction or a decimal between -1 and 1, various interesting properties arise, such as convergence.
Limit of a Sequence
The limit of a sequence is the value that the terms of the sequence approach as the sequence progresses towards infinity. It is a fundamental concept in calculus and helps us understand the behavior of sequences over the long term.
For the sequence defined by \( a_n = \frac{1}{3^n} \), we are concerned with what happens as \( n \) becomes very large. Each term gets increasingly smaller, hinting that the sequence could have a finite limit.
When finding the limit of a geometric sequence where the absolute value of the common ratio \( |r| < 1 \), the sequence approaches 0. So, for our sequence:
  • The terms become negligible as \( n \) increases.
  • The limit as \( n \to \infty \) is \( 0 \), showing the sequence's gradual tendency towards this value.
Convergent Sequence
A convergent sequence is one whose terms approach a specific value, known as the limit, as they progress to infinity. If a sequence does not approach a specific value, it is called divergent.
For instance, the sequence \( a_n = \frac{1}{3^n} \) is convergent because its terms approach 0 as \( n \) becomes larger without bound. This behavior is typical when the absolute value of the common ratio \( |r| < 1 \) for a geometric sequence, as mentioned previously.
Key aspects of convergent sequences:
  • The sequence stabilizes around a particular value.
  • Geometric sequences with a small common ratio (i.e., less than 1) tend to converge.
  • Convergent sequences are predictable and behave nicely mathematically, allowing for easier analysis and calculation of their limits.

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

Use a graphing device to determine whether the limit exists. If the limit exists, estimate its value to two decimal places. $$\lim _{x \rightarrow 2} \frac{x^{3}+6 x^{2}-5 x+1}{x^{3}-x^{2}-8 x+12}$$

Evaluate the limit if it exists. $$\lim _{x \rightarrow-4} \frac{\frac{1}{4}+\frac{1}{x}}{4+x}$$

Complete the table of values (to five decimal places), and use the table to estimate the value of the limit. $$\lim _{x \rightarrow 1} \frac{x-1}{x^{3}-1}$$ $$\begin{array}{|c|c|c|c||c|c|c|} \hline x & 0.9 & 0.99 & 0.999 & 1.001 & 1.01 & 1.1 \\ \hline f(x) & & & & & & \\ \hline \end{array}$$

Find the limit, if it exists. If the limit does not exist, explain why. $$\lim _{x \rightarrow 0^{-}}\left(\frac{1}{x}-\frac{1}{|x|}\right)$$

If the sequence is convergent, find its limit. If it is divergent, explain why. $$a_{n}=\frac{3}{n^{2}}\left[\frac{n(n+1)}{2}\right]$$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

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.