To determine whether the series

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

is convergent or divergent, we will use the Alternating Series Test (Leibniz Test), which states that a series of the form $\sum_{k=0}^{\infty} (-1)^k a_k$ converges if the following conditions are met:

1. $a_k \geq 0$ for all $k$.
2. $a_k$ is monotonically decreasing, i.e., $a_k \geq a_{k+1}$ for all $k$.
3. $\lim_{k \to \infty} a_k = 0$.

Step 1: Verify $a_k \geq 0$

Here, $a_k = \frac{1}{1 + e^k}$. Since $1 + e^k > 0$ for all $k \geq 0$, $a_k > 0$ for all $k$.

Step 2: Check if $a_k$ is monotonically decreasing

To check if $a_k$ is monotonically decreasing, we need to show that:

$a_k = \frac{1}{1 + e^k} \geq \frac{1}{1 + e^{k+1}} = a_{k+1}.$

Since $e^{k+1} > e^k$, we have $1 + e^{k+1} > 1 + e^k$, and therefore:

$\frac{1}{1 + e^k} > \frac{1}{1 + e^{k+1}},$

showing that $a_k$ is indeed decreasing.

Step 3: Evaluate $\lim_{k \to \infty} a_k$

We find:

$\lim_{k \to \infty} \frac{1}{1 + e^k} = 0,$

since $e^k$ grows without bound as $k \to \infty$, making the denominator infinitely large.

Conclusion

All conditions for the Alternating Series Test are satisfied. Therefore, the series

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

is convergent.

Do you need further explanation or have any questions?

Here are five related questions:

1. What is the difference between absolute and conditional convergence?
2. How would you apply the Ratio Test to determine the convergence of a series?
3. Can an alternating series diverge even if it satisfies some conditions of the Alternating Series Test?
4. How would the convergence of this series change if the general term was modified to $\frac{(-1)^k}{1 + k^2}$?
5. What is the behavior of the series if the alternating factor $(-1)^k$ is removed?

Tip: The Alternating Series Test only confirms conditional convergence; if you want to check for absolute convergence, you need to consider the series without the alternating sign factor.