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1b. Show that lim inf (n) = = -∞ if and only if {n} is not bounded below.

1b. Show that lim inf (n) = = -∞ if and only if {n} is not bounded below.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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### Problem 1b **Objective:** Show that the limit inferior of a sequence \( \{x_n\} \), denoted as \(\liminf(x_n) = -\infty\), if and only if the sequence \( \{x_n\} \) is not bounded below. --- **Tasks:** 1. **Argue that if \( \{x_n\} \) is not bounded below, then \(\liminf(x_n) = -\infty\).** This involves demonstrating that when there is no real number that serves as a lower bound for the sequence, the limit inferior extends to negative infinity. 2. **Show that if \(\liminf(x_n) = -\infty\), then \( \{x_n\} \) is not bounded below.** **Hint:** Suppose that \( \{x_n\} \) is bounded below, i.e., there exists an \( M \in \mathbb{R} \) such that \( M \le x_n \) for all \( n \in \mathbb{N} \). Consider the contradiction that arises when \(\liminf(x_n) = -\infty\) under this assumption. --- **Explanation:** The exercise requires establishing a bi-conditional relationship between the limit inferior reaching negative infinity and the sequence being unbounded below. You will need to consider both directions of the statement separately: - **From unbounded below to \(-\infty\):** Show that if no lower bound exists, then the eventual behavior of the sequence's limit inferior is unbounded in the negative direction. - **From \(-\infty\) to unbounded below:** Using the hint provided, attempt proof by contradiction or direct proof showing if the \(\liminf\) is \(-\infty\), the sequence must lack a lower bound. This task develops an understanding of the concept of limit inferior in relation to the boundedness properties of sequences, a fundamental aspect in analysis.
Transcribed Image Text:### Problem 1b **Objective:** Show that the limit inferior of a sequence \( \{x_n\} \), denoted as \(\liminf(x_n) = -\infty\), if and only if the sequence \( \{x_n\} \) is not bounded below. --- **Tasks:** 1. **Argue that if \( \{x_n\} \) is not bounded below, then \(\liminf(x_n) = -\infty\).** This involves demonstrating that when there is no real number that serves as a lower bound for the sequence, the limit inferior extends to negative infinity. 2. **Show that if \(\liminf(x_n) = -\infty\), then \( \{x_n\} \) is not bounded below.** **Hint:** Suppose that \( \{x_n\} \) is bounded below, i.e., there exists an \( M \in \mathbb{R} \) such that \( M \le x_n \) for all \( n \in \mathbb{N} \). Consider the contradiction that arises when \(\liminf(x_n) = -\infty\) under this assumption. --- **Explanation:** The exercise requires establishing a bi-conditional relationship between the limit inferior reaching negative infinity and the sequence being unbounded below. You will need to consider both directions of the statement separately: - **From unbounded below to \(-\infty\):** Show that if no lower bound exists, then the eventual behavior of the sequence's limit inferior is unbounded in the negative direction. - **From \(-\infty\) to unbounded below:** Using the hint provided, attempt proof by contradiction or direct proof showing if the \(\liminf\) is \(-\infty\), the sequence must lack a lower bound. This task develops an understanding of the concept of limit inferior in relation to the boundedness properties of sequences, a fundamental aspect in analysis.
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