7 A = > B=E(-1}**1 n2 n=1 N + 4n+1> n=1 ,3
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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I need help determining if these infinite series converge or diverge and why. Please show tests used or any work so I can follow along.
![The image shows two mathematical series expressions:
1. **Expression for A**:
\[
A = \sum_{n=1}^{\infty} \frac{7}{n + 4^{n+1}}
\]
This is an infinite series where each term is determined by the formula \(\frac{7}{n + 4^{n+1}}\). The series starts at \(n = 1\) and continues indefinitely.
2. **Expression for B**:
\[
B = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n^2}{\sqrt{n^3}}
\]
This is another infinite series where the terms alternate in sign due to the factor \((-1)^{n+1}\). The formula for each term is \(\frac{n^2}{\sqrt{n^3}}\). This series also begins at \(n = 1\) and continues indefinitely.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcb4f8469-eda0-4bfd-b67e-0d596825d7c6%2F2d96cdf2-530b-4664-8635-6f9a267ba8b3%2Fqx4iwms_processed.png&w=3840&q=75)
Transcribed Image Text:The image shows two mathematical series expressions:
1. **Expression for A**:
\[
A = \sum_{n=1}^{\infty} \frac{7}{n + 4^{n+1}}
\]
This is an infinite series where each term is determined by the formula \(\frac{7}{n + 4^{n+1}}\). The series starts at \(n = 1\) and continues indefinitely.
2. **Expression for B**:
\[
B = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n^2}{\sqrt{n^3}}
\]
This is another infinite series where the terms alternate in sign due to the factor \((-1)^{n+1}\). The formula for each term is \(\frac{n^2}{\sqrt{n^3}}\). This series also begins at \(n = 1\) and continues indefinitely.
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