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Let's analyze the given infinite series:

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

To determine whether the series converges or diverges, we can analyze the general behavior of the terms for large \( n \).

### Step 1: Simplify the general term for large \( n \)
For large \( n \), \( e^{2n} \) grows much faster than \( e^n \). Therefore, the denominator \( 1 + e^{2n} \) is approximately equal to \( e^{2n} \). This simplifies the general term as follows:

\[
\frac{5e^n}{1 + e^{2n}} \approx \frac{5e^n}{e^{2n}} = \frac{5}{e^n}
\]

### Step 2: Apply a known test
Now, let's analyze the behavior of \( \frac{5}{e^n} \) for large \( n \). This term behaves like a geometric series with a common ratio of \( \frac{1}{e} \), where \( e \approx 2.718 \). Since \( \frac{1}{e} < 1 \), we know that the geometric series converges.

Thus, the series behaves like a convergent geometric series for large \( n \), meaning the original series should converge as well.

### Step 3: Conclusion
The correct answer is **B**: "The series converges because it is a geometric series with \( |r| < 1 \)."

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Do you want further details on any step? Here are some related questions:

1. How does the geometric series test work for convergence?
2. What is the importance of comparing series with geometric ones?
3. Can we use other tests like the Ratio Test for this series?
4. How would the behavior change if the terms grew faster, like \( n^2 \) or \( e^{3n} \)?
5. What is the formal definition of a geometric series?

**Tip**: When analyzing series with exponential terms, it's often useful to approximate terms for large \( n \) to simplify the convergence analysis.

Does the series ∑ (5e^n / (1 + e^(2n))) from n=1 to infinity converge or diverge? Give a reason for your answer.

Learn how to determine whether the infinite series 5e^n / (1 + e^(2n)) converges or diverges using approximation and the geometric series test. Suitable for students in undergraduate or advanced high school calculus.

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