/*! 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 Which of the sequences \(\left\\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 27

Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. \(a_{n}=2+(0.1)^{n}\)

Short Answer

Expert verified
The sequence converges to 2.

Step by step solution

01

Identify Sequence Behavior

First, let's analyze the sequence given by \(a_n = 2 + (0.1)^n\). As \(n\) increases, \((0.1)^n\) becomes smaller because it's a positive number less than 1 being raised to larger powers. Therefore, \((0.1)^n\) approaches 0 as \(n\) goes to infinity.
02

Determine Convergence or Divergence

The sequence \((0.1)^n\) converges to 0 since it is a geometric sequence with a common ratio \(r = 0.1\) and \(|r| < 1\). As \((0.1)^n\) approaches 0, the entire sequence \(a_n = 2 + (0.1)^n\) approaches 2.
03

Confirm Limit of the Sequence

To find the limit of \(a_n\), consider \( \lim_{{n \to \infty}} a_n = \lim_{{n \to \infty}} \left(2 + (0.1)^n\right) = 2 + \lim_{{n \to \infty}} (0.1)^n = 2 + 0 = 2\.\) Hence, the sequence converges to 2.

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 sequences
Geometric sequences are special types of sequences where each term after the first is found by multiplying the previous term by a constant, known as the common ratio. These sequences take the form:
  • First term, often represented as \( a \)
  • Common ratio, \( r \)
  • General term for the sequence, which is: \( a, ar, ar^2, ar^3, \ldots \)
In the context of the problem, the sequence \((0.1)^n\) is geometric. Here, the first term \( a \) is 1 because \((0.1)^0 = 1\), and the common ratio \( r \) is 0.1. Since the absolute value of this common ratio is less than 1, each consecutive term gets smaller as \( n \) increases. Understanding geometric sequences is fundamental to recognizing patterns and behaviors in various mathematical contexts, particularly those involving exponential decay or growth.
limits of sequences
A limit in sequences describes the value that the terms of the sequence approach as the sequence progresses indefinitely. Simply put, when you calculate the limit of a sequence \( a_n \), you are investigating what happens to \( a_n \) as \( n \) becomes very large.
  • If a sequence \( a_n \) approaches a certain number \( L \) as \( n \) tends toward infinity, then the limit of \( a_n \) is \( L \).
  • For example, in the earlier problem, \((0.1)^n\) approaches 0 as \( n \) goes to infinity. This is because multiplying 0.1 by itself many times results in an increasingly smaller number.
  • The entire sequence \( a_n = 2 + (0.1)^n \) approaches a limit of 2 as \( n \) grows larger.
Limits are crucial in many areas of mathematics because they help us understand stable long-term behavior, even if we can't observe it directly at the beginning of a sequence.
divergence and convergence
Divergence and convergence describe whether a sequence settles into a single value (converges) or not (diverges) as it progresses.
  • A sequence converges if it approaches some finite number cut off at the end.
  • Conversely, a sequence diverges if it increases without bound, endlessly oscillates, or doesn't settle on any particular value.
  • In the example, \((0.1)^n\) converges to 0 because its terms become consecutively smaller and approach a specific value.
  • Likewise, the sequence \( a_n = 2 + (0.1)^n \) converges to 2, as the addition of \( (0.1)^n \) dwindles to making no significant impact over time.
The concept of divergence and convergence ensures clarity about the long-term behavior of sequences, allowing mathematicians to predict trends accurately and solve problems that depend on this behavior.

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

In the alternating harmonic series, suppose the goal is to arrange the terms to get a new series that converges to \(-1 / 2 .\) Start the new arrangement with the first negative term, which is \(-1 / 2\). Whenever you have a sum that is less than or equal to \(-1 / 2,\) start introducing positive terms, taken in order, until the new total is greater than \(-1 / 2 .\) Then add negative terms until the total is less than or equal to \(-1 / 2\) again. Continue this process until your partial sums have been above the target at least three times and finish at or below it. If \(s_{n}\) is the sum of the first \(n\) terms of your new series, plot the points \(\left(n, s_{n}\right)\) to illustrate how the sums are behaving.

The approximation \(e^{x}=1+x+\left(x^{2} / 2\right)\) is used when \(x\) is small. Use the Remainder Estimation Theorem to estimate the error when \(|x|<0.1\)

A triple of positive integers \(a, b,\) and \(c\) is called a Pythagorean triple if \(a^{2}+b^{2}=c^{2} .\) Let \(a\) be an odd positive integer and let $$b=\left\lfloor\frac{a^{2}}{2}\right\rfloor \text { and } c=\left\lceil\frac{a^{2}}{2}\right\rceil$$ be, respectively, the integer floor and ceiling for \(a^{2} / 2\). a. Show that \(a^{2}+b^{2}=c^{2} .\) (Hint: Let \(a=2 n+1\) and express \(b\) and \(c\) in terms of \(n .\) ) b. By direct calculation, or by appealing to the accompanying figure, find $$\lim _{a \rightarrow \infty} \frac{\left\lfloor\frac{a^{2}}{2}\right\rfloor}{\left[\frac{a^{2}}{2}\right]}$$.

Determine if the sequence is monotonic and if it is bounded. \(a_{n}=2-\frac{2}{n}-\frac{1}{2^{n}}\)

a. Let \(P\) be an approximation of \(\pi\) accurate to \(n\) decimals. Show that \(P+\sin P\) gives an approximation correct to \(3 n\) decimals. (Hint: Let \(P=\pi+x .)\) b. Try it with a calculator.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Geometry

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.