We are tasked with showing that the series

$S = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} \ln(k)}{k^2}$

converges absolutely.

Step 1: Absolute Convergence

To determine if the series converges absolutely, we consider the series formed by the absolute values of its terms:

$\sum_{k=1}^{\infty} \left| \frac{(-1)^{k+1} \ln(k)}{k^2} \right| = \sum_{k=1}^{\infty} \frac{\left| \ln(k) \right|}{k^2}$

Thus, we need to analyze the series:

$\sum_{k=1}^{\infty} \frac{\ln(k)}{k^2}$

Step 2: Comparison Test

To determine if this series converges, we can use the comparison test.

For large values of $k$, we know that $\ln(k)$ grows slower than any power of $k$. Specifically, we can compare it to the simpler series $\sum \frac{1}{k^{3/2}}$, which we know converges because the exponent $\frac{3}{2} > 1$.

Step 3: Showing the Comparison

We need to show that $\frac{\ln(k)}{k^2}$ behaves similarly to $\frac{1}{k^{3/2}}$ as $k \to \infty$. Notice that:

$\frac{\ln(k)}{k^2} \sim \frac{1}{k^{3/2}}$

for large $k$, since $\ln(k)$ grows much slower than $k^{1/2}$, and hence we expect that $\frac{\ln(k)}{k^2}$ decays faster than $\frac{1}{k^{3/2}}$.

Step 4: Limit Comparison Test

We can apply the limit comparison test with the series $\sum \frac{1}{k^{3/2}}$, which we know converges. We compute the following limit:

$\lim_{k \to \infty} \frac{\frac{\ln(k)}{k^2}}{\frac{1}{k^{3/2}}} = \lim_{k \to \infty} \frac{\ln(k)}{k^{1/2}} = 0$

Since this limit is finite and positive, the limit comparison test confirms that the series $\sum \frac{\ln(k)}{k^2}$ converges.

Step 5: Conclusion

Since $\sum_{k=1}^{\infty} \frac{\ln(k)}{k^2}$ converges, it follows that the series $\sum_{k=1}^{\infty} \left| \frac{(-1)^{k+1} \ln(k)}{k^2} \right|$ converges as well. Therefore, the series

$S = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} \ln(k)}{k^2}$

converges absolutely.

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Would you like any further details or clarification on this solution?

Here are some related questions:

1. What is the alternating series test, and how can it be applied to this series?
2. How does the comparison test work in more general cases?
3. What are other common convergence tests used for series?
4. How does the behavior of $\ln(k)$ affect the convergence of a series?
5. Can the result change if we modify the exponent of $k$ in the denominator?

Tip: When working with series, always try comparing it with a known convergent series for easier evaluation using tests like limit comparison.