4. As shown in the required reading or videos let {an) and (bn) be convergent sequences such that lim an = L₁ and lim b₁ = L2. Prove, using 71-8 N. € type methods only, that the limit operator is linear and, distributes across sums such that lim an+lim bn = L₁ + L2 21 an 21-100
4. As shown in the required reading or videos let {an) and (bn) be convergent sequences such that lim an = L₁ and lim b₁ = L2. Prove, using 71-8 N. € type methods only, that the limit operator is linear and, distributes across sums such that lim an+lim bn = L₁ + L2 21 an 21-100
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 4: Limit of the Sum of Convergent Sequences
As shown in the required reading or videos, let \(\{a_n\}\) and \(\{b_n\}\) be convergent sequences such that:
\[
\lim_{n \to \infty} a_n = L_1 \quad \text{and} \quad \lim_{n \to \infty} b_n = L_2
\]
Prove, using \(N, \epsilon\) type methods only, that the limit operator is linear and distributes across sums, such that:
\[
\lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n = L_1 + L_2
\]
### Detailed Explanation
To approach this problem, we'll use the formal definition of limits involving \(N\) and \(\epsilon\):
1. Since \(\{a_n\}\) converges to \(L_1\), for every \(\epsilon > 0\), there exists a positive integer \(N_1\) such that for all \(n \geq N_1\), \(|a_n - L_1| < \epsilon/2\).
2. Similarly, since \(\{b_n\}\) converges to \(L_2\), for every \(\epsilon > 0\), there exists a positive integer \(N_2\) such that for all \(n \geq N_2\), \(|b_n - L_2| < \epsilon/2\).
Given these definitions, our goal is to show that \(\lim_{n \to \infty} (a_n + b_n) = L_1 + L_2\).
#### Step-by-Step Proof:
1. **Combine the Inequalities:**
We aim to combine the inequalities for \(|a_n - L_1|\) and \(|b_n - L_2|\). Since \(\epsilon\) is arbitrary, we choose \(\epsilon/2\) to ensure we can add the two sequences while controlling both within a total of \(\epsilon\).
2. **Find a Common \(N\):**
Let \(N = \max(N_1, N_2)\). This selection ensures both inequalities \(|a_n - L_1| < \epsilon/2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcdbdf4c8-5459-45de-99f8-cd8d71031beb%2F0901a40e-7ca8-4e60-9573-d2a6ff6c2a5d%2Fajkjdwe_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 4: Limit of the Sum of Convergent Sequences
As shown in the required reading or videos, let \(\{a_n\}\) and \(\{b_n\}\) be convergent sequences such that:
\[
\lim_{n \to \infty} a_n = L_1 \quad \text{and} \quad \lim_{n \to \infty} b_n = L_2
\]
Prove, using \(N, \epsilon\) type methods only, that the limit operator is linear and distributes across sums, such that:
\[
\lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n = L_1 + L_2
\]
### Detailed Explanation
To approach this problem, we'll use the formal definition of limits involving \(N\) and \(\epsilon\):
1. Since \(\{a_n\}\) converges to \(L_1\), for every \(\epsilon > 0\), there exists a positive integer \(N_1\) such that for all \(n \geq N_1\), \(|a_n - L_1| < \epsilon/2\).
2. Similarly, since \(\{b_n\}\) converges to \(L_2\), for every \(\epsilon > 0\), there exists a positive integer \(N_2\) such that for all \(n \geq N_2\), \(|b_n - L_2| < \epsilon/2\).
Given these definitions, our goal is to show that \(\lim_{n \to \infty} (a_n + b_n) = L_1 + L_2\).
#### Step-by-Step Proof:
1. **Combine the Inequalities:**
We aim to combine the inequalities for \(|a_n - L_1|\) and \(|b_n - L_2|\). Since \(\epsilon\) is arbitrary, we choose \(\epsilon/2\) to ensure we can add the two sequences while controlling both within a total of \(\epsilon\).
2. **Find a Common \(N\):**
Let \(N = \max(N_1, N_2)\). This selection ensures both inequalities \(|a_n - L_1| < \epsilon/2
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