Use the Limit Comparison Test to determine the convergence or divergence of the series. lim n18 n = 1 8 9 +1 8 + 1 9n + 1 + 1 =L>0
Use the Limit Comparison Test to determine the convergence or divergence of the series. lim n18 n = 1 8 9 +1 8 + 1 9n + 1 + 1 =L>0
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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Question
![Use the Limit Comparison Test to determine the convergence or divergence of the series.
\[
\sum_{n=1}^{\infty} \frac{8^n + 1}{9^n + 1}
\]
\[
\lim_{{n \to \infty}} \frac{\frac{8^n + 1}{9^n + 1}}{\text{[comparison series]}} = L > 0
\]
There is a box suggesting that you need to choose a suitable comparison series to complete the Limit Comparison Test.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf8dd5bb-1368-4a10-9929-9294802d0b74%2F7492b473-6ee1-419f-8ee0-8468ef54e57c%2F7fm0z8m_processed.png&w=3840&q=75)
Transcribed Image Text:Use the Limit Comparison Test to determine the convergence or divergence of the series.
\[
\sum_{n=1}^{\infty} \frac{8^n + 1}{9^n + 1}
\]
\[
\lim_{{n \to \infty}} \frac{\frac{8^n + 1}{9^n + 1}}{\text{[comparison series]}} = L > 0
\]
There is a box suggesting that you need to choose a suitable comparison series to complete the Limit Comparison Test.
![**Use the Limit Comparison Test to determine the convergence or divergence of the series.**
\[
\sum_{n=1}^{\infty} \frac{1}{n^4(n^4 + 8)}
\]
\[
\lim_{n \to \infty} \frac{1}{n^4(n^4 + 8)} \div \boxed{\phantom{\text{comparison series}}} = L > 0
\]
**Explanation:**
- The problem involves using the Limit Comparison Test to determine whether the given infinite series converges or diverges.
- The expression to be tested is \(\sum_{n=1}^{\infty} \frac{1}{n^4(n^4 + 8)}\).
- The Limit Comparison Test involves comparing this series with another series (represented by a blank box) known to converge or diverge. The limit \(L\) (which must be greater than zero) will indicate convergence or divergence.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf8dd5bb-1368-4a10-9929-9294802d0b74%2F7492b473-6ee1-419f-8ee0-8468ef54e57c%2F1mgnf4l_processed.png&w=3840&q=75)
Transcribed Image Text:**Use the Limit Comparison Test to determine the convergence or divergence of the series.**
\[
\sum_{n=1}^{\infty} \frac{1}{n^4(n^4 + 8)}
\]
\[
\lim_{n \to \infty} \frac{1}{n^4(n^4 + 8)} \div \boxed{\phantom{\text{comparison series}}} = L > 0
\]
**Explanation:**
- The problem involves using the Limit Comparison Test to determine whether the given infinite series converges or diverges.
- The expression to be tested is \(\sum_{n=1}^{\infty} \frac{1}{n^4(n^4 + 8)}\).
- The Limit Comparison Test involves comparing this series with another series (represented by a blank box) known to converge or diverge. The limit \(L\) (which must be greater than zero) will indicate convergence or divergence.
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