For which real r> 0 does the series n=1 converge, and for which real r> 0 does the series diverge? (5+3r)n + √n (8+r)n + n²
For which real r> 0 does the series n=1 converge, and for which real r> 0 does the series diverge? (5+3r)n + √n (8+r)n + n²
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 Statement
For which real \( r > 0 \) does the series
\[
\sum_{n=1}^{\infty} \frac{(5 + 3r)^n + \sqrt{n}}{(8 + r)^n + n^2}
\]
converge, and for which real \( r > 0 \) does the series diverge?
---
### Explanation
This problem involves determining the values of \( r \) for which the given infinite series converges or diverges. The series in question is:
\[
\sum_{n=1}^{\infty} \frac{(5 + 3r)^n + \sqrt{n}}{(8 + r)^n + n^2}
\]
We need to analyze the behavior of the terms as \( n \) approaches infinity, which will depend on the relative growth rates of \((5 + 3r)^n\) and \((8 + r)^n\).
### Analysis
1. **Numerator**: \((5 + 3r)^n + \sqrt{n}\)
- For large \( n \), the term \((5 + 3r)^n\) tends to dominate over \(\sqrt{n}\) if \((5 + 3r)^n\) grows exponentially, much faster than any polynomial term like \(\sqrt{n}\).
2. **Denominator**: \((8 + r)^n + n^2\)
- Similarly, \((8 + r)^n\) tends to dominate over \(n^2\) for large \( n \).
### Dominant Term Comparison
For large \( n \), compare the dominant exponential terms:
\[
\frac{(5 + 3r)^n}{(8 + r)^n}
\]
- If \((5 + 3r) < (8 + r)\), the base of the exponential term in the numerator is smaller than that in the denominator, causing the ratio to tend to zero as \( n \) increases. Hence, the series converges.
- If \((5 + 3r) \ge (8 + r)\), the ratio does not tend to zero, which leads to divergence.
### Conclusion
The series converges for \( 5 + 3r < 8 + r \), simplifying to:
\[
2r <](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F77ccc228-61ae-45bc-bff3-36e89ce5abe9%2F1752f356-a811-4b13-84f4-398bcd751532%2F4fudi8e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
For which real \( r > 0 \) does the series
\[
\sum_{n=1}^{\infty} \frac{(5 + 3r)^n + \sqrt{n}}{(8 + r)^n + n^2}
\]
converge, and for which real \( r > 0 \) does the series diverge?
---
### Explanation
This problem involves determining the values of \( r \) for which the given infinite series converges or diverges. The series in question is:
\[
\sum_{n=1}^{\infty} \frac{(5 + 3r)^n + \sqrt{n}}{(8 + r)^n + n^2}
\]
We need to analyze the behavior of the terms as \( n \) approaches infinity, which will depend on the relative growth rates of \((5 + 3r)^n\) and \((8 + r)^n\).
### Analysis
1. **Numerator**: \((5 + 3r)^n + \sqrt{n}\)
- For large \( n \), the term \((5 + 3r)^n\) tends to dominate over \(\sqrt{n}\) if \((5 + 3r)^n\) grows exponentially, much faster than any polynomial term like \(\sqrt{n}\).
2. **Denominator**: \((8 + r)^n + n^2\)
- Similarly, \((8 + r)^n\) tends to dominate over \(n^2\) for large \( n \).
### Dominant Term Comparison
For large \( n \), compare the dominant exponential terms:
\[
\frac{(5 + 3r)^n}{(8 + r)^n}
\]
- If \((5 + 3r) < (8 + r)\), the base of the exponential term in the numerator is smaller than that in the denominator, causing the ratio to tend to zero as \( n \) increases. Hence, the series converges.
- If \((5 + 3r) \ge (8 + r)\), the ratio does not tend to zero, which leads to divergence.
### Conclusion
The series converges for \( 5 + 3r < 8 + r \), simplifying to:
\[
2r <
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