Chapter 10: Problem 5
The sequence \(\left\\{r^{n}\right\\}\) converges if and only if (A) \(|r|^{<} 1\) (B) \(|r| \leqq 1\) (C) \(-1^{<} r \leqq 1\) (D) \(0^{<} r^{<} 1\)
Short Answer
Expert verified
The correct answer is (A) \(|r| < 1\).
Step by step solution
01
Understand Sequence Convergence
A sequence \(\{a_n\}\) converges if there exists a real number \(L\) such that for every positive number \(\varepsilon\), there is a positive integer \(N\) where for all \(n \geq N\), \(|a_n - L| < \varepsilon\). In simpler terms, the terms of the sequence get arbitrarily close to \(L\) as \(n\) becomes larger.
02
Identify Sequence Behavior of \(r^n\)
The sequence \(r^n\) depends primarily on \(|r|\). If \(|r| < 1\), the terms \(r^n\) diminish to 0 as \(n\) increases. If \(|r| > 1\), the terms \(r^n\) grow larger or oscillate between large magnitudes. If \(|r| = 1\), the limit depends on the exact value of \(r\): if \(r = 1\), \(r^n = 1\) (convergent), but if \(r = -1\), \(r^n\) oscillates (divergent).
03
Analyze the Choices
Based on our analysis of \(r\):- (A) \(|r| < 1\) implies convergence to 0.- (B) \(|r| \leq 1\) is not sufficient; \(|r| = 1\) can oscillate (e.g., \(r = -1\)).- (C) \(-1 < r \leq 1\) still allows for oscillation.- (D) \(0 < r < 1\) implies convergence to 0 (but excludes 0, which should actually be included for convergence to 0).
04
Determine the Correct Choice
Upon evaluating all conditions, (A) \(|r| < 1\) accurately describes when the sequence \(r^n\) converges, because it ensures that the terms\(r^n\) approach 0 regardless of \(n\). This matches our understanding of convergence toward a limit.
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit of a Sequence
The concept of a limit is fundamental in understanding sequence convergence. When we talk about the limit of a sequence like \( \{a_n\} \), we mean a certain value, let's call it \( L \), that the terms of the sequence get arbitrarily close to as the sequence progresses. This is often written as \( \lim_{{n \to \infty}} a_n = L \). For a sequence to have a limit, and hence converge, a specific condition must be met: for any small, positive number, known as \( \varepsilon \), there must be a stage in the sequence past which all terms are within \( \varepsilon \) of \( L \). This stage is represented by a natural number \( N \), where for all \( n \geq N \), the expression \(|a_n - L| < \varepsilon\) holds true. This implies that, however tiny the gap we consider, there will be a point in the sequence at which the terms will fall inside that gap after a while, indicating that they are converging towards the limit \( L \). The result is that essentially, the terms of the sequence get closer and closer to \( L \), even though they might never exactly reach \( L \).
Geometric Sequences
Geometric sequences are a particular type of sequence where each term is obtained by multiplying the previous term by a constant, known as the common ratio \( r \). If the first term of a geometric sequence is \( a \), then the terms are \( a, ar, ar^2, ar^3, \ldots \). The sequence can be represented as \( \{ r^n \} \) when the first term \( a \) is 1. The behavior of a geometric sequence strongly depends on the value of \(|r|\), the absolute value of the common ratio.
- If \(|r| < 1\), the sequence approaches zero as \( n \) becomes very large. This results in a convergent sequence.- For \(|r| > 1\), the terms become larger, tending towards infinity, or oscillate wildly if \( r \) is negative.- When \(|r| = 1\), the sequence might either stay constant (if \( r = 1 \)) or oscillate between values (if \( r = -1 \)).Understanding these behaviors helps determine whether such sequences converge and to what value they may converge, such as zero in the case where \(|r| < 1\).
- If \(|r| < 1\), the sequence approaches zero as \( n \) becomes very large. This results in a convergent sequence.- For \(|r| > 1\), the terms become larger, tending towards infinity, or oscillate wildly if \( r \) is negative.- When \(|r| = 1\), the sequence might either stay constant (if \( r = 1 \)) or oscillate between values (if \( r = -1 \)).Understanding these behaviors helps determine whether such sequences converge and to what value they may converge, such as zero in the case where \(|r| < 1\).
Convergence Criteria
To ascertain the convergence of a sequence, one must apply specific criteria. The convergence criteria for a sequence concern limits and absolute values, especially in the realm of geometric sequences. The sequence \( r^n \) will converge depending on the absolute value of \( r \): - **Convergence to Zero**: If \(|r| < 1\), the sequence \( r^n \) approaches zero because each successive multiplication by \( r \) (a fraction) reduces the magnitude of the term.- **Divergence**: If \(|r| > 1\), the sequence diverges as each multiplication by \( r \) (a number greater than one) increases the term's magnitude.- **Oscillation and Divergence**: When \(|r| = 1\), it's a special case. The sequence might oscillate, such as when \( r = -1 \), alternating between 1 and -1, and thus doesn't settle at a single value (except in the trivial case where all terms are 1).Applying these criteria, it's straightforward to identify the conditions under which \( r^n \) converges. The correct criterion for convergence in geometric sequences is \(|r| < 1\), indicating guaranteed convergence to zero, simplifying the analysis of sequence behavior considerably.