The given series is:

$1 - \ln(2) + \frac{(\ln(2))^2}{2!} - \frac{(\ln(2))^3}{3!} + \frac{(\ln(2))^4}{4!} - \dots$

This series resembles the Maclaurin series (Taylor series expansion around $x = 0$) of the exponential function $e^x$, which is:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$

Here, the signs alternate between positive and negative, suggesting it matches the series expansion of $e^{-x}$, which is:

$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \dots$

Substitution of $x = \ln(2)$:

Setting $x = \ln(2)$ into $e^{-x}$, we get:

$e^{-\ln(2)} = \frac{1}{e^{\ln(2)}} = \frac{1}{2}.$

Conclusion:

The sum of the given series is:

$\boxed{\frac{1}{2}}$

Would you like further clarification on any step?

Related Questions:

1. What is the Maclaurin series for $e^x$ and $e^{-x}$?
2. How do you determine the sum of an infinite geometric series?
3. What is the role of factorials in Taylor or Maclaurin series?
4. Why does substituting $x = \ln(2)$ simplify the exponential function?
5. How do you verify the convergence of a series?

Tip:

Always compare a given series to known series (e.g., exponential, trigonometric, logarithmic) for quick identification and summation!