Problem 16
Question
Let \(\left\\{x_{n}\right\\}\) be a sequence. Suppose there are two convergent subsequences \(\left\\{x_{n_{i}}\right\\}\) and \(\left\\{x_{m_{i}}\right\\} .\) Suppose $$\lim _{i \rightarrow \infty} x_{n_{i}}=a \quad \text { and } \quad \lim _{i \rightarrow \infty} x_{m_{i}}=b,$$ where \(a \neq b\). Prove that \(\left\\{x_{n}\right\\}\) is not convergent, without using Proposition 2.1.17.
Step-by-Step Solution
Verified Answer
The sequence \( \{x_n\} \) is not convergent because it has two subsequences converging to different limits \( a \) and \( b \), where \( a \neq b \).
1Step 1: Understand the Assumptions
We are given a sequence \( \{x_n\} \) and two convergent subsequences within it, \( \{x_{n_i}\} \) and \( \{x_{m_i}\} \). These subsequences converge to different limits, \( a \) and \( b \) respectively, and it is stated that \( a eq b \). We need to show that the original sequence \( \{x_n\} \) is not convergent.
2Step 2: Recall the Definition of Convergence
A sequence \( \{x_n\} \) is said to converge to a limit \( L \) if for every \( \epsilon > 0 \), there exists a natural number \( N \) such that for all \( n > N \), \( |x_n - L| < \epsilon \). This means all terms of the sequence eventually get arbitrarily close to \( L \).
3Step 3: Use the Contrapositive
In contrast, if \( \{x_n\} \) does not converge, it means for some \( \epsilon > 0 \), no matter how large \( N \) is, it's not true that all \( n > N \) satisfy \( |x_n - L| < \epsilon \).
4Step 4: Analyze the Behavior of Subsequence Limits
Suppose \( \{x_n\} \) converged to some limit \( L \). Then, every subsequence of \( \{x_n\} \) must also converge to \( L \) (a property of convergent sequences). However, we have two subsequences converging to different limits \( a \) and \( b \) with \( a eq b \). This shows a contradiction because a sequence cannot converge to more than one limit.
5Step 5: Conclude the Contradiction
Since a single sequence \( \{x_n\} \) cannot simultaneously have two different limits, the assumption that \( \{x_n\} \) is convergent must be false. Therefore, \( \{x_n\} \) is not convergent.
Key Concepts
SubsequencesConvergent SequencesLimit of a Sequence
Subsequences
A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. Let's consider a sequence \( \{ x_n \} \). To form a subsequence, you might choose terms indexed by increasing integers like \( n_1, n_2, n_3, \.\.\. \). Thus, a subsequence would be \( \{ x_{n_i} \} \).
Subsequences retain the inherent properties of the original sequence but can behave differently, especially when it comes to convergence.
It's possible, as seen in this exercise, to have subsequences that converge to different limits while originating from the same sequence.
Understanding subsequences is crucial:
Subsequences retain the inherent properties of the original sequence but can behave differently, especially when it comes to convergence.
It's possible, as seen in this exercise, to have subsequences that converge to different limits while originating from the same sequence.
Understanding subsequences is crucial:
- They allow us to isolate or highlight certain behaviors within a sequence.
- They provide a means to test and explore sequence convergence.
- They help us understand the overall nature of the original sequence.
Convergent Sequences
A convergent sequence is one where the terms tend to get closer and closer to a specific value, known as the limit, as the sequence progresses. For a sequence \( \{ x_n \} \) to converge to a limit \( L \), each term in the sequence should eventually be within any arbitrarily small distance from \( L \).
This means we can find a natural number \( N \) such that for every term in the sequence beyond this point, the distance between \( x_n \) and \( L \) is less than any small number \( \epsilon \) we choose.
Important aspects include:
This foundational property underscores why \( \{ x_n \} \) isn’t convergent, as a convergent sequence’s behavior must be consistent for every part or version of it, including subsequences.
This means we can find a natural number \( N \) such that for every term in the sequence beyond this point, the distance between \( x_n \) and \( L \) is less than any small number \( \epsilon \) we choose.
Important aspects include:
- The uniqueness of the limit: A sequence can converge to at most one limit.
- Persistence: Once close to the limit, terms stay close; they don’t wander off significantly.
- Every subsequence of a convergent sequence shares the same limit.
This foundational property underscores why \( \{ x_n \} \) isn’t convergent, as a convergent sequence’s behavior must be consistent for every part or version of it, including subsequences.
Limit of a Sequence
The limit of a sequence \( \{ x_n \} \) is the value that the sequence terms approach closer and closer as \( n \) increases indefinitely. If such a limit \( L \) exists, we say the sequence converges to \( L \), denoted mathematically as \( \lim_{n \to \infty} x_n = L \).
The concept of a limit is central to understanding sequence convergence and has several key aspects:
This supports the reasoning for why non-uniform limits among subsequences indicate the original sequence doesn't converge, since convergence implies a singular, predominant directional limit for all components.
The concept of a limit is central to understanding sequence convergence and has several key aspects:
- The limit must be reached for every ensuing part of the sequence beyond a certain point.
- The sequence can get as close as desired to its limit by taking sufficiently large terms.
- No matter how small a distance we specify, the sequence terms will eventually all lie within that distance from the limit.
This supports the reasoning for why non-uniform limits among subsequences indicate the original sequence doesn't converge, since convergence implies a singular, predominant directional limit for all components.
Other exercises in this chapter
Problem 16
Prove that \(\left\\{(n !)^{1 / n}\right\\}\) is unbounded. Hint: Show that \(\left\\{\frac{C^{n}}{n !}\right\\}\) converges to zero for any \(C>0\).
View solution Problem 16
Suppose \(\left\\{x_{n}\right\\}\) is a bounded sequence, \(a_{n}:=\sup \left\\{x_{k}: k \geq n\right\\}\) as before. Suppose that for some \(\ell \in \mathbb{N
View solution Problem 17
Suppose \(\left\\{x_{n}\right\\}\) is a sequence, and \(a_{n}:=\sup \left\\{x_{k}: k \geq n\right\\}\) and \(b_{n}:=\sup \left\\{x_{k}: k \geq n\right\\}\) as b
View solution Problem 17
Prove Abel's theorem: Theorem. Suppose \(\sum x_{n}\) is a series whose partial sums are a bounded sequence, \(\left\\{\lambda_{n}\right\\}\) is a sequence with
View solution