Determine convergence/ divergence. If result is convergence, find the li
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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Determine convergence/ divergence. If result is convergence, find the limit.
![### Sequences and Series: Example Summations
- **Part d**:
The given series is:
\[
\sum_{n=1}^{\infty} \frac{4}{16n^2 - 8n - 3}
\]
This represents an infinite series where each term is given by the fraction \(\frac{4}{16n^2 - 8n - 3}\). The denominator is a quadratic expression in terms of \(n\).
- **Part e**:
The given series is:
\[
\sum_{n=1}^{\infty} \frac{e^{\pi n}}{\pi^{n e}}
\]
This represents an infinite series where each term is given by \(\frac{e^{\pi n}}{\pi^{n e}}\). The term involves exponential functions of both \(\pi\) and \(e\).
- **Part f**:
The given series is:
\[
\sum_{n=1}^{\infty} \left( \frac{1}{2^{1/n}} - \frac{1}{2^{1/(n+1)}} \right)
\]
This represents an infinite series where each term is the difference of two fractions. Each fraction is given as \( \frac{1}{2^{1/n}} \) and \( \frac{1}{2^{1/(n+1)}} \).
In these series:
- \(\sum\) denotes the sum.
- \(n=1\) to \(\infty\): The index \(n\) starts from 1 and goes to infinity.
- \(e\): Euler's number, approximately equal to 2.71828.
- \(\pi\): Pi, approximately equal to 3.14159.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc787ae5f-ecbe-4215-b4d0-f569392eb050%2Fa60d7e6f-5311-454f-8f51-6ef9198c13f8%2Fhoh0i3_processed.png&w=3840&q=75)
Transcribed Image Text:### Sequences and Series: Example Summations
- **Part d**:
The given series is:
\[
\sum_{n=1}^{\infty} \frac{4}{16n^2 - 8n - 3}
\]
This represents an infinite series where each term is given by the fraction \(\frac{4}{16n^2 - 8n - 3}\). The denominator is a quadratic expression in terms of \(n\).
- **Part e**:
The given series is:
\[
\sum_{n=1}^{\infty} \frac{e^{\pi n}}{\pi^{n e}}
\]
This represents an infinite series where each term is given by \(\frac{e^{\pi n}}{\pi^{n e}}\). The term involves exponential functions of both \(\pi\) and \(e\).
- **Part f**:
The given series is:
\[
\sum_{n=1}^{\infty} \left( \frac{1}{2^{1/n}} - \frac{1}{2^{1/(n+1)}} \right)
\]
This represents an infinite series where each term is the difference of two fractions. Each fraction is given as \( \frac{1}{2^{1/n}} \) and \( \frac{1}{2^{1/(n+1)}} \).
In these series:
- \(\sum\) denotes the sum.
- \(n=1\) to \(\infty\): The index \(n\) starts from 1 and goes to infinity.
- \(e\): Euler's number, approximately equal to 2.71828.
- \(\pi\): Pi, approximately equal to 3.14159.
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