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.

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.