Complete the theorem in two ways.A) Prove {xn} converges by modifying the proof for increasing sequences that is presented in the text.B) Prove that {xn} converges by considering the sequence {-xn} and using the result for increasing sequences The text provides an analysis of two mathematical sequences, explaining their properties and convergence using mathematical induction and other techniques.### Explanation of the Sequences1. **Sequence \( a_n \):** - The sequence \( a_n \) is defined by the sum: \[ a_n = \sum_{k=1}^{n} \frac{1}{k \cdot 2^k} \leq \sum_{k=1}^{n} \frac{1}{2^k} = 1 - \frac{1}{2^n} < 1 \] - Every term \( a_{n+1} \) is obtained from \( a_n \) by adding a positive number, indicating the sequence is increasing. - The sequence is bounded by 1 and is positive. - By Theorem 2.10, since \( a_n \) is bounded and increasing, it is convergent.2. **Sequence \( b_n \):** - The sequence \( b_n \) is defined recursively with \( b_1 = 3 \) and: \[ b_{n+1} = \frac{b_n}{2} + \frac{3}{b_n} \] - The first four terms are given as 3, 2.5, 2.45, and 2.4495, indicating a strictly decreasing sequence. - Verification that \( b_{n+1} - b_n < 0 \) involves showing: \[ \frac{b_n}{2} + \frac{3}{b_n} - b_n = \frac{6 - b_n^2}{2b_n} \] - This inequality simplifies to \( b_n^2 > 6 \) for all \( n \), which is proved by induction. - The sequence is bounded below by \( \sqrt{6} \) and converges by Theorem 2.10.### Convergence and Limits- **Exact Value of Limits:** - The limit of a sequence can be approximated, but finding the exact value can be tricky without known limits. - Numerical analysis helps in approximation, though exact determination methods are preferred for recursive sequences.- **Finding Limit \( L \) of \( b_n \):** - Let \( L \) be the limit of \( b **Theorem 2.10**: A monotone sequence converges if and only if it is bounded.**Proof**: Since every convergent sequence is bounded (see Theorem 2.5), it remains to prove that every bounded monotone sequence converges. Let \(\{x_n\}\) be an increasing sequence that is bounded. Since the sequence \(\{x_n\}\) is bounded, the Completeness Axiom guarantees that the number \(\beta = \sup\{x_n : n \in \mathbb{Z}^+\}\) exists. Let \(\epsilon > 0\). Since \(\beta - \epsilon\) is not an upper bound of the set \(\{x_n : n \in \mathbb{Z}^+\}\), there exists a positive integer \(N\) such that \(x_N > \beta - \epsilon\). Now use the fact that \(\{x_n\}\) is increasing to conclude that\[\beta - \epsilon < x_N \leq x_n \leq \beta\]for all \(n \geq N\) (see Figure 2.3). It follows that \(|x_n - \beta| < \epsilon\) for all \(n \geq N\). Therefore, the sequence \(\{x_n\}\) converges to \(\beta\). A proof that bounded decreasing sequences converge will be left as an exercise. This completes the proof.To illustrate how this theorem is used to prove that a sequence converges, consider the sequence \(\{a_n\}\), where \(a_n = \sum_{k=1}^{n} 1/k2^k\) for each positive integer \(n\). The first six terms of this sequence are\[\frac{1}{2}, \quad \frac{5}{8}, \quad \frac{2}{3}, \quad \frac{131}{192}, \quad \frac{661}{960}, \quad \text{and} \quad \frac{1327}{1920}.\]

(A) Suppose xn s a monotone convergent sequence. Since every convergent sequence is bounded,…

Answered: A monotone sequence converges if and…

Math

Advanced Math

A monotone sequence converges if and only if it is bounded.

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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Question

Complete the theorem in two ways. A) Prove {xn} converges by modifying the proof for increasing sequences that is presented in the text. B) Prove that {xn} converges by considering the sequence {-xn} and using the result for increasing sequences

The text provides an analysis of two mathematical sequences, explaining their properties and convergence using mathematical induction and other techniques.

### Explanation of the Sequences

1. **Sequence \( a_n \):**

- The sequence \( a_n \) is defined by the sum:
\[
a_n = \sum_{k=1}^{n} \frac{1}{k \cdot 2^k} \leq \sum_{k=1}^{n} \frac{1}{2^k} = 1 - \frac{1}{2^n} < 1
\]
- Every term \( a_{n+1} \) is obtained from \( a_n \) by adding a positive number, indicating the sequence is increasing.
- The sequence is bounded by 1 and is positive.
- By Theorem 2.10, since \( a_n \) is bounded and increasing, it is convergent.

2. **Sequence \( b_n \):**

- The sequence \( b_n \) is defined recursively with \( b_1 = 3 \) and:
\[
b_{n+1} = \frac{b_n}{2} + \frac{3}{b_n}
\]
- The first four terms are given as 3, 2.5, 2.45, and 2.4495, indicating a strictly decreasing sequence.
- Verification that \( b_{n+1} - b_n < 0 \) involves showing:
\[
\frac{b_n}{2} + \frac{3}{b_n} - b_n = \frac{6 - b_n^2}{2b_n}
\]
- This inequality simplifies to \( b_n^2 > 6 \) for all \( n \), which is proved by induction.
- The sequence is bounded below by \( \sqrt{6} \) and converges by Theorem 2.10.

### Convergence and Limits

- **Exact Value of Limits:**
- The limit of a sequence can be approximated, but finding the exact value can be tricky without known limits.
- Numerical analysis helps in approximation, though exact determination methods are preferred for recursive sequences.

- **Finding Limit \( L \) of \( b_n \):**
- Let \( L \) be the limit of \( b

**Theorem 2.10**: A monotone sequence converges if and only if it is bounded.

**Proof**: Since every convergent sequence is bounded (see Theorem 2.5), it remains to prove that every bounded monotone sequence converges. Let \(\{x_n\}\) be an increasing sequence that is bounded. Since the sequence \(\{x_n\}\) is bounded, the Completeness Axiom guarantees that the number \(\beta = \sup\{x_n : n \in \mathbb{Z}^+\}\) exists. Let \(\epsilon > 0\). Since \(\beta - \epsilon\) is not an upper bound of the set \(\{x_n : n \in \mathbb{Z}^+\}\), there exists a positive integer \(N\) such that \(x_N > \beta - \epsilon\). Now use the fact that \(\{x_n\}\) is increasing to conclude that

\[
\beta - \epsilon < x_N \leq x_n \leq \beta
\]

for all \(n \geq N\) (see Figure 2.3). It follows that \(|x_n - \beta| < \epsilon\) for all \(n \geq N\). Therefore, the sequence \(\{x_n\}\) converges to \(\beta\). A proof that bounded decreasing sequences converge will be left as an exercise. This completes the proof.

To illustrate how this theorem is used to prove that a sequence converges, consider the sequence \(\{a_n\}\), where \(a_n = \sum_{k=1}^{n} 1/k2^k\) for each positive integer \(n\). The first six terms of this sequence are

\[
\frac{1}{2}, \quad \frac{5}{8}, \quad \frac{2}{3}, \quad \frac{131}{192}, \quad \frac{661}{960}, \quad \text{and} \quad \frac{1327}{1920}.
\]

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