### Series Analysis Consider the series given by: \[ \sum_{n=1}^{\infty} \frac{(-1)^n n^3}{7^n} \] We need to determine whether this series converges absolutely, converges conditionally, or diverges. Additionally, we seek to find the limit if it does converge. #### Steps to Analyze the Series 1. **Check for Absolute Convergence** - The series converges absolutely if the series of absolute values converges. - Consider the series: \[ \sum_{n=1}^{\infty} \left| \frac{(-1)^n n^3}{7^n} \right| = \sum_{n=1}^{\infty} \frac{n^3}{7^n} \] 2. **Comparison with a Known Convergent Series** - Analyze the terms of the series \(\frac{n^3}{7^n}\). - Use the Ratio Test: \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\) \[ \lim_{n \to \infty} \left| \frac{\frac{(n+1)^3}{7^{n+1}}}{\frac{n^3}{7^n}} \right| = \lim_{n \to \infty} \left| \frac{(n+1)^3}{7 \cdot n^3} \right| = \frac{1}{7} \lim_{n \to \infty} \left| \frac{(n+1)^3}{n^3} \right| = \frac{1}{7} \lim_{n \to \infty} \left| 1 + \frac{3}{n} + \frac{3}{n^2} + \frac{1}{n^3} \right| = \frac{1}{7} \] - Since \(\frac{1}{7} < 1\), by the Ratio Test, the series \(\sum_{n=1}^{\infty} \frac{n^3}{7^n}\) converges. - Therefore, the original series also converges absolutely by the Comparison Test. 3. **Conclusion**
### Series Analysis Consider the series given by: \[ \sum_{n=1}^{\infty} \frac{(-1)^n n^3}{7^n} \] We need to determine whether this series converges absolutely, converges conditionally, or diverges. Additionally, we seek to find the limit if it does converge. #### Steps to Analyze the Series 1. **Check for Absolute Convergence** - The series converges absolutely if the series of absolute values converges. - Consider the series: \[ \sum_{n=1}^{\infty} \left| \frac{(-1)^n n^3}{7^n} \right| = \sum_{n=1}^{\infty} \frac{n^3}{7^n} \] 2. **Comparison with a Known Convergent Series** - Analyze the terms of the series \(\frac{n^3}{7^n}\). - Use the Ratio Test: \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\) \[ \lim_{n \to \infty} \left| \frac{\frac{(n+1)^3}{7^{n+1}}}{\frac{n^3}{7^n}} \right| = \lim_{n \to \infty} \left| \frac{(n+1)^3}{7 \cdot n^3} \right| = \frac{1}{7} \lim_{n \to \infty} \left| \frac{(n+1)^3}{n^3} \right| = \frac{1}{7} \lim_{n \to \infty} \left| 1 + \frac{3}{n} + \frac{3}{n^2} + \frac{1}{n^3} \right| = \frac{1}{7} \] - Since \(\frac{1}{7} < 1\), by the Ratio Test, the series \(\sum_{n=1}^{\infty} \frac{n^3}{7^n}\) converges. - Therefore, the original series also converges absolutely by the Comparison Test. 3. **Conclusion**
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Series Analysis
Consider the series given by:
\[ \sum_{n=1}^{\infty} \frac{(-1)^n n^3}{7^n} \]
We need to determine whether this series converges absolutely, converges conditionally, or diverges. Additionally, we seek to find the limit if it does converge.
#### Steps to Analyze the Series
1. **Check for Absolute Convergence**
- The series converges absolutely if the series of absolute values converges.
- Consider the series: \[ \sum_{n=1}^{\infty} \left| \frac{(-1)^n n^3}{7^n} \right| = \sum_{n=1}^{\infty} \frac{n^3}{7^n} \]
2. **Comparison with a Known Convergent Series**
- Analyze the terms of the series \(\frac{n^3}{7^n}\).
- Use the Ratio Test: \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\)
\[
\lim_{n \to \infty} \left| \frac{\frac{(n+1)^3}{7^{n+1}}}{\frac{n^3}{7^n}} \right| = \lim_{n \to \infty} \left| \frac{(n+1)^3}{7 \cdot n^3} \right| = \frac{1}{7} \lim_{n \to \infty} \left| \frac{(n+1)^3}{n^3} \right| = \frac{1}{7} \lim_{n \to \infty} \left| 1 + \frac{3}{n} + \frac{3}{n^2} + \frac{1}{n^3} \right| = \frac{1}{7}
\]
- Since \(\frac{1}{7} < 1\), by the Ratio Test, the series \(\sum_{n=1}^{\infty} \frac{n^3}{7^n}\) converges.
- Therefore, the original series also converges absolutely by the Comparison Test.
3. **Conclusion**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad630951-4c60-4c80-8c8a-9fae0bc83c36%2F5fb23e08-6edb-4ee6-acd1-3dec428a2630%2F03z0a0j.jpeg&w=3840&q=75)
Transcribed Image Text:### Series Analysis
Consider the series given by:
\[ \sum_{n=1}^{\infty} \frac{(-1)^n n^3}{7^n} \]
We need to determine whether this series converges absolutely, converges conditionally, or diverges. Additionally, we seek to find the limit if it does converge.
#### Steps to Analyze the Series
1. **Check for Absolute Convergence**
- The series converges absolutely if the series of absolute values converges.
- Consider the series: \[ \sum_{n=1}^{\infty} \left| \frac{(-1)^n n^3}{7^n} \right| = \sum_{n=1}^{\infty} \frac{n^3}{7^n} \]
2. **Comparison with a Known Convergent Series**
- Analyze the terms of the series \(\frac{n^3}{7^n}\).
- Use the Ratio Test: \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\)
\[
\lim_{n \to \infty} \left| \frac{\frac{(n+1)^3}{7^{n+1}}}{\frac{n^3}{7^n}} \right| = \lim_{n \to \infty} \left| \frac{(n+1)^3}{7 \cdot n^3} \right| = \frac{1}{7} \lim_{n \to \infty} \left| \frac{(n+1)^3}{n^3} \right| = \frac{1}{7} \lim_{n \to \infty} \left| 1 + \frac{3}{n} + \frac{3}{n^2} + \frac{1}{n^3} \right| = \frac{1}{7}
\]
- Since \(\frac{1}{7} < 1\), by the Ratio Test, the series \(\sum_{n=1}^{\infty} \frac{n^3}{7^n}\) converges.
- Therefore, the original series also converges absolutely by the Comparison Test.
3. **Conclusion**
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