Theorem 8. If lim a, = a and lim b, = b, then lim (a,b,) = ab. %3D
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
100%
#68
![**Convergence of Sequences and Series**
**Armed with this bound \( B \), we can add on one more inequality to the above scrapwork to get:**
\[ |a_n \cdot b_n - a \cdot b| = |a_n \cdot b_n - a \cdot b + a_n \cdot b - a_n \cdot b| \]
\[ \leq |a_n \cdot b_n - a_n \cdot b| + |a_n \cdot b - a \cdot b| \]
\[ = |a_n| \cdot |b_n - b| + |b| \cdot |a_n - a| \]
\[ < B |b_n - b| + (|b| + 1) |a_n - a| \]
*At this point, we should be able to make the last line of this less than \(\epsilon\).*
**END OF SCRAPWORK**
**Problem 68.** Prove Theorem 8. \( \diamond \)
**Corollary 1.** (Corollary to Theorem 8) If \(\lim_{n \to \infty} a_n = a\) and \(c \in \mathbb{R}\), then \(\lim_{n \to \infty} c \cdot a_n = c \cdot a\).
**Problem 69.** Prove the above corollary to Theorem 8. \( \diamond \)
*^1 Actually, this is a dangerous habit to fall into even in convergence proofs.*
---
**Convergence of Sequences and Series**
**Theorem 9.** Suppose \(\lim_{n \to \infty} a_n = a\) and \(\lim_{n \to \infty} b_n = b\). Also suppose \(b \neq 0\) and \(b_n \neq 0\), \(\forall n\). Then \(\lim_{n \to \infty} \left( \frac{a_n}{b_n} \right) = \frac{a}{b}\). \( \blacktriangle \)
**SCRAPWORK:** To prove this, let's look at the special case of trying to prove \(\lim_{n \to \infty} \left( \frac{b_n](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc074fc7e-4e48-46f1-bfc4-aee6ba4e6262%2Fbf5b980b-f875-4351-899f-e0a11e400f5b%2Fz6y368b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Convergence of Sequences and Series**
**Armed with this bound \( B \), we can add on one more inequality to the above scrapwork to get:**
\[ |a_n \cdot b_n - a \cdot b| = |a_n \cdot b_n - a \cdot b + a_n \cdot b - a_n \cdot b| \]
\[ \leq |a_n \cdot b_n - a_n \cdot b| + |a_n \cdot b - a \cdot b| \]
\[ = |a_n| \cdot |b_n - b| + |b| \cdot |a_n - a| \]
\[ < B |b_n - b| + (|b| + 1) |a_n - a| \]
*At this point, we should be able to make the last line of this less than \(\epsilon\).*
**END OF SCRAPWORK**
**Problem 68.** Prove Theorem 8. \( \diamond \)
**Corollary 1.** (Corollary to Theorem 8) If \(\lim_{n \to \infty} a_n = a\) and \(c \in \mathbb{R}\), then \(\lim_{n \to \infty} c \cdot a_n = c \cdot a\).
**Problem 69.** Prove the above corollary to Theorem 8. \( \diamond \)
*^1 Actually, this is a dangerous habit to fall into even in convergence proofs.*
---
**Convergence of Sequences and Series**
**Theorem 9.** Suppose \(\lim_{n \to \infty} a_n = a\) and \(\lim_{n \to \infty} b_n = b\). Also suppose \(b \neq 0\) and \(b_n \neq 0\), \(\forall n\). Then \(\lim_{n \to \infty} \left( \frac{a_n}{b_n} \right) = \frac{a}{b}\). \( \blacktriangle \)
**SCRAPWORK:** To prove this, let's look at the special case of trying to prove \(\lim_{n \to \infty} \left( \frac{b_n
![Transcription for Educational Website:
---
To understand the convergence of sequences and series, let's consider the problem of making a particular expression smaller as desired. We have the following inequality:
\[ |(a_n + b_n) - (a + b)| = |(a_n - a) + (b_n - b)| \leq |a_n - a| + |b_n - b| \]
This is derived using the triangle inequality. To keep the entire expression less than ε, we aim to make each part less than ε/2. Fortunately, the definition of limits for sequences allows us to choose N large enough so that these differences are arbitrarily small:
1. As \(\lim_{n \to \infty} a_n = a\), there exists some \(N_1\) such that for \(n > N_1\), \(|a_n - a| < \frac{\epsilon}{2}\).
2. Similarly, \(\lim_{n \to \infty} b_n = b\) leads to the existence of an \(N_2\) where for \(n > N_2\), \(|b_n - b| < \frac{\epsilon}{2}\).
Thus, we set \(N = \max(N_1, N_2)\) to satisfy both conditions.
**Problem 66:** Prove Theorem 7.
**Theorem 8:** If \(\lim_{n \to \infty} a_n = a\) and \(\lim_{n \to \infty} b_n = b\), then \(\lim_{n \to \infty} (a_n b_n) = ab\).
**SCRAPWORK:** Given ε > 0, choose N such that for \(n > N\), \(|a_n b_n - ab| < \epsilon\). In analysis, we often "uncancel" by adding and subtracting a judiciously chosen term, allowing us to employ methods like the Reverse Triangle Inequality (Problem 55).
For instance, consider:
\[ |a_n b_n - ab| = |a_n b_n - a_n b + a_n b - ab| \]
\[ \leq |a_n b_n - a_n b| + |a_n b - ab| \]
\[ = |a_n| |b_n - b| + |b| |a_n -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc074fc7e-4e48-46f1-bfc4-aee6ba4e6262%2Fbf5b980b-f875-4351-899f-e0a11e400f5b%2Flv6gmnm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Transcription for Educational Website:
---
To understand the convergence of sequences and series, let's consider the problem of making a particular expression smaller as desired. We have the following inequality:
\[ |(a_n + b_n) - (a + b)| = |(a_n - a) + (b_n - b)| \leq |a_n - a| + |b_n - b| \]
This is derived using the triangle inequality. To keep the entire expression less than ε, we aim to make each part less than ε/2. Fortunately, the definition of limits for sequences allows us to choose N large enough so that these differences are arbitrarily small:
1. As \(\lim_{n \to \infty} a_n = a\), there exists some \(N_1\) such that for \(n > N_1\), \(|a_n - a| < \frac{\epsilon}{2}\).
2. Similarly, \(\lim_{n \to \infty} b_n = b\) leads to the existence of an \(N_2\) where for \(n > N_2\), \(|b_n - b| < \frac{\epsilon}{2}\).
Thus, we set \(N = \max(N_1, N_2)\) to satisfy both conditions.
**Problem 66:** Prove Theorem 7.
**Theorem 8:** If \(\lim_{n \to \infty} a_n = a\) and \(\lim_{n \to \infty} b_n = b\), then \(\lim_{n \to \infty} (a_n b_n) = ab\).
**SCRAPWORK:** Given ε > 0, choose N such that for \(n > N\), \(|a_n b_n - ab| < \epsilon\). In analysis, we often "uncancel" by adding and subtracting a judiciously chosen term, allowing us to employ methods like the Reverse Triangle Inequality (Problem 55).
For instance, consider:
\[ |a_n b_n - ab| = |a_n b_n - a_n b + a_n b - ab| \]
\[ \leq |a_n b_n - a_n b| + |a_n b - ab| \]
\[ = |a_n| |b_n - b| + |b| |a_n -
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
