Chapter 10: Problem 9
Determine whether the given sequence converges. $$ \left\\{\frac{n^{2}-1}{2 n}\right\\} $$
Short Answer
Expert verified
The sequence does not converge.
Step by step solution
01
Understanding the Sequence
First, observe that the sequence is given by \( a_n = \frac{n^2 - 1}{2n} \). We want to determine if this sequence converges as \( n \to \infty \).
02
Simplifying the Expression
Rewrite the sequence term as \( a_n = \frac{n^2}{2n} - \frac{1}{2n} = \frac{n}{2} - \frac{1}{2n} \). This breakdown simplifies the expression into two components that are easier to analyze.
03
Analyze Each Component
Consider each part separately: \( \frac{n}{2} \) and \( \frac{1}{2n} \). As \( n \to \infty \), \( \frac{1}{2n} \to 0 \) because the denominator grows without bound. However, \( \frac{n}{2} \to \infty \) as n grows.
04
Conclusion on Convergence
The expression \( \frac{n}{2} - \frac{1}{2n} \to \infty - 0 = \infty \). Since it approaches infinity, the sequence does not converge.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit of a sequence
When we discuss the **limit of a sequence**, we are trying to understand what happens to the values of a sequence as we increase the index value, often denoted by \( n \). A sequence converges if the terms become closer and closer to a specific number as \( n \) tends towards infinity. More formally, a sequence \( \{a_n\} \) is said to converge to \( L \) if for every positive number \( \epsilon \), there exists a positive integer \( N \) such that for all \( n > N \), the absolute difference \( |a_n - L| < \epsilon \). This means that beyond a certain index, all terms of the sequence are arbitrarily close to \( L \).
For the sequence given by \( a_n = \frac{n^2 - 1}{2n} \), evaluating its limit helps determine whether it converges or diverges. Upon simplification, you find that the dominant term as \( n \to \infty \) is \( \frac{n}{2} \), which grows without bound, indicating that the limit does not exist in the finite realm, and thus the sequence does not converge.
For the sequence given by \( a_n = \frac{n^2 - 1}{2n} \), evaluating its limit helps determine whether it converges or diverges. Upon simplification, you find that the dominant term as \( n \to \infty \) is \( \frac{n}{2} \), which grows without bound, indicating that the limit does not exist in the finite realm, and thus the sequence does not converge.
Asymptotic behavior
Understanding the **asymptotic behavior** of a sequence helps in comprehending how the sequences behave as they stretch towards infinity. It's a way to describe the limiting behavior of a sequence without necessarily finding an exact limit.
In the exercise, to identify the asymptotic behavior, the sequence \( \frac{n^2 - 1}{2n} \) was expressed as \( \frac{n}{2} - \frac{1}{2n} \). Here, \( \frac{n}{2} \) suggests that as \( n \) increases, the sequence largely behaves like \( n/2 \), growing larger and larger. The term \( \frac{1}{2n} \) becomes negligible as \( n \) grows very large, having less and less impact on the total value of the sequence. Thus, asymptotically, the sequence behaves like \( \frac{n}{2} \).
This insight makes it clear that as \( n \to \infty \), the sequence's terms also move towards infinity, confirming that it does not settle at a fixed value, which reflects divergence.
In the exercise, to identify the asymptotic behavior, the sequence \( \frac{n^2 - 1}{2n} \) was expressed as \( \frac{n}{2} - \frac{1}{2n} \). Here, \( \frac{n}{2} \) suggests that as \( n \) increases, the sequence largely behaves like \( n/2 \), growing larger and larger. The term \( \frac{1}{2n} \) becomes negligible as \( n \) grows very large, having less and less impact on the total value of the sequence. Thus, asymptotically, the sequence behaves like \( \frac{n}{2} \).
This insight makes it clear that as \( n \to \infty \), the sequence's terms also move towards infinity, confirming that it does not settle at a fixed value, which reflects divergence.
Infinite sequences
An **infinite sequence** refers to a list of numbers in a specific order that continues indefinitely. In mathematical terms, it means there is no last term. Some infinite sequences may "settle down" to a particular number, implying they converge, while others do not and are said to diverge.
The infinite sequence in question, \( \left\{\frac{n^2 - 1}{2n}\right\} \), does not become stable as \( n \to \infty \). Instead, its terms grow bigger and approach infinity. This behavior characterizes diverging infinite sequences.
Each new term of this sequence is derived by plugging in values of \( n \) that continue to increase without bound. Because the main term \( \frac{n}{2} \) dominates, effectively leading to an infinite progression, the sequence lacks any fixed limit. This is crucial for differentiating between sequences that may have trivial patterns but never actually converge to a finite point.
The infinite sequence in question, \( \left\{\frac{n^2 - 1}{2n}\right\} \), does not become stable as \( n \to \infty \). Instead, its terms grow bigger and approach infinity. This behavior characterizes diverging infinite sequences.
Each new term of this sequence is derived by plugging in values of \( n \) that continue to increase without bound. Because the main term \( \frac{n}{2} \) dominates, effectively leading to an infinite progression, the sequence lacks any fixed limit. This is crucial for differentiating between sequences that may have trivial patterns but never actually converge to a finite point.