(c) (d) In n (−1)” +00 n=1 2n³ 3 n + +1 +00 n=1 n
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
![**Problem Statement:**
Write out the first five terms of the sequence, determine whether the sequence converges, and if so, find its limit.
---
**Sequence (c):**
\[
\left\{ \frac{\ln n}{n} \right\}_{n=1}^{\infty}
\]
**Steps to Solve:**
1. **Calculate the first five terms:**
- When \( n = 1 \), \(\frac{\ln 1}{1} = 0\)
- When \( n = 2 \), \(\frac{\ln 2}{2} \approx 0.347\)
- When \( n = 3 \), \(\frac{\ln 3}{3} \approx 0.366\)
- When \( n = 4 \), \(\frac{\ln 4}{4} = 0.346\)
- When \( n = 5 \), \(\frac{\ln 5}{5} \approx 0.321\)
2. **Convergence Analysis:**
- As \( n \to \infty \), \(\frac{\ln n}{n} \to 0\)
- **Limit:** The sequence converges to 0.
---
**Sequence (d):**
\[
\left\{ \frac{(-1)^n \cdot 2n^3}{n^3 + 1} \right\}_{n=1}^{\infty}
\]
**Steps to Solve:**
1. **Calculate the first five terms:**
- When \( n = 1 \), \(\frac{(-1)^1 \cdot 2 \cdot 1^3}{1^3 + 1} = -1\)
- When \( n = 2 \), \(\frac{(-1)^2 \cdot 2 \cdot 8}{8 + 1} \approx 1.778\)
- When \( n = 3 \), \(\frac{(-1)^3 \cdot 2 \cdot 27}{27 + 1} \approx -1.929\)
- When \( n = 4 \), \(\frac{(-1)^4 \cdot 2 \cdot 64}{64 + 1} \approx 1.969\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc57f7e0d-797b-4b4c-aff5-3e2cb2f516c9%2Faeb7d896-630b-4775-abd8-1c08ba675660%2Fbmjr7b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Write out the first five terms of the sequence, determine whether the sequence converges, and if so, find its limit.
---
**Sequence (c):**
\[
\left\{ \frac{\ln n}{n} \right\}_{n=1}^{\infty}
\]
**Steps to Solve:**
1. **Calculate the first five terms:**
- When \( n = 1 \), \(\frac{\ln 1}{1} = 0\)
- When \( n = 2 \), \(\frac{\ln 2}{2} \approx 0.347\)
- When \( n = 3 \), \(\frac{\ln 3}{3} \approx 0.366\)
- When \( n = 4 \), \(\frac{\ln 4}{4} = 0.346\)
- When \( n = 5 \), \(\frac{\ln 5}{5} \approx 0.321\)
2. **Convergence Analysis:**
- As \( n \to \infty \), \(\frac{\ln n}{n} \to 0\)
- **Limit:** The sequence converges to 0.
---
**Sequence (d):**
\[
\left\{ \frac{(-1)^n \cdot 2n^3}{n^3 + 1} \right\}_{n=1}^{\infty}
\]
**Steps to Solve:**
1. **Calculate the first five terms:**
- When \( n = 1 \), \(\frac{(-1)^1 \cdot 2 \cdot 1^3}{1^3 + 1} = -1\)
- When \( n = 2 \), \(\frac{(-1)^2 \cdot 2 \cdot 8}{8 + 1} \approx 1.778\)
- When \( n = 3 \), \(\frac{(-1)^3 \cdot 2 \cdot 27}{27 + 1} \approx -1.929\)
- When \( n = 4 \), \(\frac{(-1)^4 \cdot 2 \cdot 64}{64 + 1} \approx 1.969\)
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