Problem 13 - Basic Analysis: Introduction to Real Analysis

Question

(Easy): Prove the following stronger version of Lemma 2.2.12, the ratio test. Suppose \(\left\\{x_{n}\right\\}\) is a sequence such that \(x_{n} \neq 0\) for all \(n\). a) Prove that if there exists an \(r<1\) and \(M \in \mathbb{N}\) such that for all \(n \geq M\) we have $$ \frac{\left|x_{n+1}\right|}{\left|x_{n}\right|} \leq r $$ then \(\left\\{x_{n}\right\\}\) converges to \(0 .\) b) Prove that if there exists an \(r>1\) and \(M \in \mathbb{N}\) such that for all \(n \geq M\) we have $$ \frac{\left|x_{n+1}\right|}{\left|x_{n}\right|} \geq r $$ then \(\left\\{x_{n}\right\\}\) is unbounded.

Step-by-Step Solution

Answer

(a) Sequence converges to 0; (b) Sequence is unbounded.

Step 1: Understand the Hypothesis for Part (a)

We are given that there exists an \(r<1\) and \(M \in \mathbb{N}\) such that for all \(n \geq M\), the condition \(\frac{\left|x_{n+1}\right|}{\left|x_{n}\right|} \leq r\) holds. Our goal is to show that the sequence \(\{x_n\}\) converges to 0.

Step 2: Express Terms of the Sequence

For \(n \geq M\), using the given condition, we can express the terms of the sequence as \(|x_{n+1}| \leq r \cdot |x_n|\). Repeated application shows \(|x_{n+k}| \leq r^k \cdot |x_n|\).

Step 3: Show the Sequence Converges to Zero

Since \(r<1\), as \(k\) becomes large, \(r^k\) approaches zero. Thus, \(|x_{n+k}| \leq r^k \cdot |x_n|\) implies \(|x_{n+k}| \to 0\) as \(k \to \infty\). Therefore, the sequence converges to 0.

Step 4: Verification for Part (a)

We have shown that the sequence converges to zero because \(|x_{n+k}|\) becomes arbitrarily small for large \(k\) since \(r^k\) shrinks.

Step 4: Understand the Hypothesis for Part (b)

For part (b), we need to show that if there exists an \(r>1\) and \(M \in \mathbb{N}\) such that for all \(n \geq M\), \(\frac{\left|x_{n+1}\right|}{\left|x_{n}\right|} \geq r\), then \(\{x_n\}\) is unbounded.

Step 5: Derive the Terms for Part (b)

According to the condition, for \(n \geq M\), we have \(|x_{n+1}| \geq r \cdot |x_n|\). Repeated application gives \(|x_{n+k}| \geq r^k \cdot |x_n|\).

Step 6: Show the Sequence is Unbounded

Since \(r>1\), as \(k\) increases, \(r^k\) grows without bound. Thus, \(|x_{n+k}| \geq r^k \cdot |x_n|\) becomes arbitrarily large, proving that \(\{x_n\}\) is unbounded.

Step 8: Verification for Part (b)

We have demonstrated that the sequence is unbounded because \(|x_{n+k}|\) can become infinitely large for sufficiently large \(k\).

Key Concepts

Sequence Convergence

In real analysis, the idea of sequence convergence is fundamental. It describes the behavior of sequences as they approach a specific value as the sequence progresses. To say a sequence \( \{a_n\} \) converges to a limit \( L \), it means:

• The absolute difference \( |a_n - L| \) becomes arbitrarily small as \( n \) becomes very large.
• Formally, for any given small \( \varepsilon > 0 \), there is an \( N \) such that for all \( n \geq N \), \( |a_n - L| < \varepsilon \).

In the context of the exercise, it means that the sequence \( \{x_n\} \) approaches zero as \( n \to \infty \). Part (a) of the step-by-step solution demonstrates this. With \( r < 1 \), each subsequent term \( |x_{n+k}| = r^k \cdot |x_n| \) becomes smaller, eventually nearing zero. Thus, given these conditions, \( \{x_n\} \) is said to converge to zero.

Ratio Test

The Ratio Test is a useful tool in determining the convergence of a series. It generally applies to infinite series, but its principles can be used for sequences as well.

• Consider the ratio \( \frac{|a_{n+1}|}{|a_n|} \) for the terms of a series or sequence.
• For a sequence to converge, it's often useful to find an \( r < 1 \) such that the inequality \( \frac{|x_{n+1}|}{|x_n|} \leq r \) holds eventually for all terms. The terms essentially decrease rapidly.

In the performed exercise, Part (a) used the ratio test to show convergence. Here, the repeated application of the ratio \( r < 1 \), as explained, ensures that \( |x_{n+k}| \) shrinks to zero. In Part (b) of the exercise, if \( r > 1 \), it implies divergence or unboundedness. This means the terms of the sequence grow larger without constraint, not approaching any finite limit, which confirms the sequence is unbounded.

Boundedness

Boundedness in sequences or series is about whether the sequence remains within a finite range as it progresses.

• A sequence \( \{a_n\} \) is bounded if there exists some number \( M \) such that \( |a_n| \leq M \) for all terms \( n \).
• Bounded sequences do not surpass certain fixed heights (or depths), essentially remaining within a given interval.

The task from Part (b) addresses unboundedness, a contrasting concept. If a sequence can be shown to eventually have terms grow larger than any number, infinitely, then the sequence is unbounded. With \( r > 1 \), as the steps show, \(|x_{n+k}| \geq r^k \cdot |x_n|\) implies that the sequence cannot be contained within any finite limit. Consequently, as \( r^k \) becomes infinitely large, the sequence \( \{x_n\} \) does not have a finite bound, demonstrating its growth as unbounded.