we series Σ n=2 4n² + 1 1 - n² is absolutely convergent is conditionally convergent, but not absolutely convergent is divergent
we series Σ n=2 4n² + 1 1 - n² is absolutely convergent is conditionally convergent, but not absolutely convergent is divergent
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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![**Convergence of an Infinite Series**
Consider the series:
\[
\sum_{n=2}^{\infty} \left( \frac{4n^2 + 1}{1 - n^2} \right)^n
\]
Determine the nature of convergence for this series:
- **Option A**: The series is absolutely convergent.
- **Option B**: The series is conditionally convergent, but not absolutely convergent.
- **Option C**: The series is divergent.
To solve this, explore the behavior of the terms and apply convergence tests where necessary. Consider absolute convergence first; if not feasible, investigate conditional convergence or divergence.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0a03efe2-d931-4461-b3ce-5e694888ce83%2Fcd8230bc-4ab1-49b6-b5d2-10ad8b3b56d7%2Flq43bfm_processed.png&w=3840&q=75)
Transcribed Image Text:**Convergence of an Infinite Series**
Consider the series:
\[
\sum_{n=2}^{\infty} \left( \frac{4n^2 + 1}{1 - n^2} \right)^n
\]
Determine the nature of convergence for this series:
- **Option A**: The series is absolutely convergent.
- **Option B**: The series is conditionally convergent, but not absolutely convergent.
- **Option C**: The series is divergent.
To solve this, explore the behavior of the terms and apply convergence tests where necessary. Consider absolute convergence first; if not feasible, investigate conditional convergence or divergence.
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