1 for convergence, you 2k5 +9k8 will use the Limit Comparison Test, comparing it to the p 1 series 9k² To test the series k=1 k=1 k=1 where p Now by the limit comparison test, the series ∞ 1 2/5 +9k8 = diverges converges O
1 for convergence, you 2k5 +9k8 will use the Limit Comparison Test, comparing it to the p 1 series 9k² To test the series k=1 k=1 k=1 where p Now by the limit comparison test, the series ∞ 1 2/5 +9k8 = diverges converges O
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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![To test the series
\[
\sum_{k=1}^{\infty} \frac{1}{2k^5 + 9k^8}
\]
for convergence, you will use the Limit Comparison Test, comparing it to the p-series
\[
\sum_{k=1}^{\infty} \frac{1}{9k^p}
\]
where \( p = \) [blank box].
Now by the limit comparison test, the series
\[
\sum_{k=1}^{\infty} \frac{1}{2k^5 + 9k^8}
\]
diverges [checkbox] converges [checkbox].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F54cb8c7a-f5c8-4f70-beb9-4c302d85da57%2Fc2153d89-095e-4196-bce7-47d3547494ec%2Fydnd069_processed.jpeg&w=3840&q=75)
Transcribed Image Text:To test the series
\[
\sum_{k=1}^{\infty} \frac{1}{2k^5 + 9k^8}
\]
for convergence, you will use the Limit Comparison Test, comparing it to the p-series
\[
\sum_{k=1}^{\infty} \frac{1}{9k^p}
\]
where \( p = \) [blank box].
Now by the limit comparison test, the series
\[
\sum_{k=1}^{\infty} \frac{1}{2k^5 + 9k^8}
\]
diverges [checkbox] converges [checkbox].
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