The problem involves the infinite series:

$\sum_{n=1}^\infty (-1)^{n-1} \frac{\ln(\ln(n))}{\sqrt{n}}$

We will analyze whether this series converges or diverges and, if it converges, determine its behavior.

Step 1: Convergence of Alternating Series (Leibniz Test)

For an alternating series of the form $\sum_{n=1}^\infty (-1)^n a_n$, the Leibniz test states that the series converges if:

1. $a_n$ is positive, decreasing, and
2. $\lim_{n \to \infty} a_n = 0$.

Here, $a_n = \frac{\ln(\ln(n))}{\sqrt{n}}$.

Step 2: Check the terms $a_n$

1. Positivity:

   • For $n > 1$, both $\ln(n) > 0$ and $\ln(\ln(n)) > 0$, so $a_n > 0$.

2. Limit as $n \to \infty$:

   • Compute $\lim_{n \to \infty} \frac{\ln(\ln(n))}{\sqrt{n}}$.
   • The square root $\sqrt{n}$ grows much faster than $\ln(\ln(n))$, so: $\lim_{n \to \infty} \frac{\ln(\ln(n))}{\sqrt{n}} = 0.$

3. Decreasing:

   • To check if $a_n$ is decreasing, observe the derivative of $f(n) = \frac{\ln(\ln(n))}{\sqrt{n}}$. The numerator $\ln(\ln(n))$ grows slower than the denominator $\sqrt{n}$, so $a_n$ decreases for large $n$.

Thus, $a_n$ satisfies all conditions of the Leibniz test.

Step 3: Conclusion

By the Leibniz test, the series converges.

Would you like a more detailed analysis or help with evaluating the series further? Here are some related questions to explore:

1. What is the exact convergence criterion for alternating series?
2. How does the growth rate of $\ln(\ln(n))$ compare to other logarithmic terms?
3. What happens if we remove the alternating factor $(-1)^{n-1}$?
4. How do other convergence tests (e.g., ratio test, root test) apply here?
5. What is the physical or practical interpretation of such alternating series?

Tip: For alternating series, always verify positivity, limit behavior, and monotonicity of the terms.