https://file.aiquickdraw.com/mathbot/file/1730174427518lirn87ye.jpg

To determine whether the series

\[
\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{2}{4n + 5e^n}
\]

converges or diverges using the Limit Comparison Test, we will proceed as follows:

1. **Choose a Comparison Series:**  
   Since the denominator \(4n + 5e^n\) grows very quickly due to the exponential term \(e^n\), the term \(\frac{2}{5e^n}\) will eventually dominate for large \(n\). Thus, we can compare \(a_n\) to a similar term:

   \[
   b_n = \frac{2}{5e^n} = \frac{2}{5} \cdot e^{-n}.
   \]

   This series resembles a geometric series, \( \sum ar^{n-1} \), where \( a = \frac{2}{5} \) and \( r = \frac{1}{e} \).

2. **Apply the Limit Comparison Test:**  
   To use the test, we find

   \[
   \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{2}{4n + 5e^n}}{\frac{2}{5e^n}} = \lim_{n \to \infty} \frac{5e^n}{4n + 5e^n}.
   \]

   Dividing the numerator and denominator by \(e^n\), we get:

   \[
   \lim_{n \to \infty} \frac{5}{\frac{4n}{e^n} + 5}.
   \]

   Since \(\frac{4n}{e^n} \to 0\) as \(n \to \infty\), the limit becomes:

   \[
   \lim_{n \to \infty} \frac{5}{0 + 5} = 1.
   \]

   Since the limit is a positive finite number (1), the Limit Comparison Test applies.

3. **Determine Convergence of \(b_n\):**  
   The series \( \sum b_n = \sum \frac{2}{5} e^{-n} \) is a geometric series with ratio \( r = \frac{1}{e} < 1 \). Therefore, \( \sum b_n \) converges.

4. **Conclusion:**  
   By the Limit Comparison Test, since \( \sum b_n \) converges, the series \( \sum a_n \) also converges.

### Final Answers:

- **Comparison series \( b_n \):** \(\frac{2}{5} e^{-n}\) with \( a = \frac{2}{5} \) and \( r = \frac{1}{e} \).
- **Limit:** \(\lim_{n \to \infty} \frac{a_n}{b_n} = 1\).
- **Convergence:** The series \( \sum a_n \) **converges** by the Limit Comparison Test.

Would you like more details on any part of this explanation?

---

### Follow-up Questions

1. What is the general form of the Limit Comparison Test?
2. How does the exponential function \( e^n \) influence the growth rate of sequences?
3. What properties define a geometric series, and why is convergence determined by \( |r| < 1 \)?
4. Could a different comparison series be used here, and would it change the conclusion?
5. What other convergence tests could be applied to a series with exponential terms?

### Tip
When choosing a comparison series, look for terms in \( a_n \) that dominate as \( n \to \infty \). This will often lead you to an effective comparison series.

Use the Limit Comparison Test to determine whether the series \( \sum_{n=1}^{\infty} \frac{2}{4n + 5e^n} \) converges or diverges.

Explore the convergence of the series \( \sum \frac{2}{4n + 5e^n} \) using the Limit Comparison Test. This solution walks through selecting a comparison series, calculating limits, and determining convergence based on a geometric series. Ideal for calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation