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n³+2n²+3n-5 Does the series Σ=1 n6+3 converge or diverge? By what test?

n³+2n²+3n-5 Does the series Σ=1 n6+3 converge or diverge? By what test?

Advanced Engineering Mathematics
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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**7.** Does the series \( \sum_{n=1}^{\infty} \frac{n^3 + 2n^2 + 3n - 5}{n^6 + 3} \) converge or diverge? By what test? **Explanation:** To determine whether the infinite series converges or diverges, we need to use appropriate tests from calculus or real analysis. In this case, one of the suitable tests to apply might be the Limit Comparison Test or the use of asymptotic behavior analysis (comparing the leading terms of the numerator and denominator for large \( n \)). **Analysis:** 1. **Leading Term Comparison**: - Consider the leading terms of the numerator and the denominator as \( n \) approaches infinity. - The leading term in the numerator \( n^3 + 2n^2 + 3n - 5 \) is \( n^3 \). - The leading term in the denominator \( n^6 + 3 \) is \( n^6 \). 2. **Forming a Simple Comparison Series**: - \( \frac{n^3}{n^6} = \frac{1}{n^3} \). 3. **Testing with a P-Series**: - The series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges if \( p > 1 \) and diverges if \( p \leq 1 \). - Here, \( p = 3 \), so \( \sum_{n=1}^{\infty} \frac{1}{n^3} \) converges. 4. **Applying the Limit Comparison Test**: - Compute the limit: \[ \lim_{{n \to \infty}} \frac{\frac{n^3 + 2n^2 + 3n - 5}{n^6 + 3}}{\frac{1}{n^3}} = \lim_{{n \to \infty}} \frac{n^3(n^3 + 2n^2 + 3n - 5)}{n^6 + 3} = \lim_{{n \to \infty}} \frac{n^6 + 2n^5 + 3n^4 - 5n
Transcribed Image Text:**7.** Does the series \( \sum_{n=1}^{\infty} \frac{n^3 + 2n^2 + 3n - 5}{n^6 + 3} \) converge or diverge? By what test? **Explanation:** To determine whether the infinite series converges or diverges, we need to use appropriate tests from calculus or real analysis. In this case, one of the suitable tests to apply might be the Limit Comparison Test or the use of asymptotic behavior analysis (comparing the leading terms of the numerator and denominator for large \( n \)). **Analysis:** 1. **Leading Term Comparison**: - Consider the leading terms of the numerator and the denominator as \( n \) approaches infinity. - The leading term in the numerator \( n^3 + 2n^2 + 3n - 5 \) is \( n^3 \). - The leading term in the denominator \( n^6 + 3 \) is \( n^6 \). 2. **Forming a Simple Comparison Series**: - \( \frac{n^3}{n^6} = \frac{1}{n^3} \). 3. **Testing with a P-Series**: - The series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges if \( p > 1 \) and diverges if \( p \leq 1 \). - Here, \( p = 3 \), so \( \sum_{n=1}^{\infty} \frac{1}{n^3} \) converges. 4. **Applying the Limit Comparison Test**: - Compute the limit: \[ \lim_{{n \to \infty}} \frac{\frac{n^3 + 2n^2 + 3n - 5}{n^6 + 3}}{\frac{1}{n^3}} = \lim_{{n \to \infty}} \frac{n^3(n^3 + 2n^2 + 3n - 5)}{n^6 + 3} = \lim_{{n \to \infty}} \frac{n^6 + 2n^5 + 3n^4 - 5n
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