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The given series is:

\[
\sum_{k=0}^{\infty} \frac{(-1)^k}{1 + e^{-k}}
\]

### Step 1: Analyze the General Term

The general term of the series is:

\[
a_k = \frac{(-1)^k}{1 + e^{-k}}
\]

Let's examine the denominator more closely:
\[
1 + e^{-k} = 1 + \frac{1}{e^k}
\]

As \( k \to \infty \), the term \( e^{-k} \to 0 \), so the denominator approaches \( 1 \). Hence, for large \( k \), the general term behaves like:
\[
a_k \approx \frac{(-1)^k}{1} = (-1)^k
\]

### Step 2: Apply Alternating Series Test (Leibniz Test)

Since the series alternates in sign due to the factor \( (-1)^k \), we can consider the Alternating Series Test. This test requires two conditions:

1. The terms \( b_k = \frac{1}{1 + e^{-k}} \) must decrease in magnitude.
2. \( \lim_{k \to \infty} b_k = 0 \).

Let's check both:

1. **Monotonicity**: The function \( f(k) = \frac{1}{1 + e^{-k}} \) is a decreasing function because \( e^{-k} \) decreases as \( k \) increases, making the denominator larger and the fraction smaller.
   
2. **Limit**: As \( k \to \infty \),
\[
\lim_{k \to \infty} \frac{1}{1 + e^{-k}} = \frac{1}{1 + 0} = 1
\]

### Step 3: Conclusion on Convergence

The Alternating Series Test fails because \( \lim_{k \to \infty} b_k = 1 \neq 0 \), which violates the second condition of the test. Since the general term does not tend to zero, the series **diverges**.

### Final Answer:
The series diverges because the terms do not approach zero.

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Would you like any further clarification or a deeper breakdown? 

Here are 5 related questions:
1. What is the Alternating Series Test?
2. How do you determine if a series converges using the limit comparison test?
3. What happens if the general term of a series does not approach zero?
4. Can a series with terms involving \( e^{-k} \) ever converge?
5. How does the behavior of the exponential function influence series convergence?

**Tip:** Always check if the general term of a series tends to zero. If it doesn't, the series diverges.

Determine if the series ∑ (-1)^k / (1 + e^(-k)) is convergent or divergent. Make sure to justify your answer.

This page explains how to determine if the series ∑ (-1)^k / (1 + e^(-k)) is convergent or divergent. It covers alternating series, exponential functions, and the conditions of the Alternating Series Test.

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