Determine the equivalent capacitance between points A and B for the...

Let the equivalent capacitance between points $A$ and $B$ be $C_{eq}$.

Because the ladder is infinite, the circuit to the right of the first $2\ \mu\text{F}$ capacitor is identical to the entire circuit and thus has an equivalent capacitance of $C_{eq}$.

The circuit can be simplified to a $1\ \mu\text{F}$ capacitor in parallel with the series combination of a $2\ \mu\text{F}$ capacitor and $C_{eq}$.

The equivalent capacitance is given by:
$C_{eq} = 1 + \dfrac{2 \cdot C_{eq}}{2 + C_{eq}}$

Multiplying both sides by $(2 + C_{eq})$:
$C_{eq}(2 + C_{eq}) = 1(2 + C_{eq}) + 2C_{eq}$

$2C_{eq} + C_{eq}^2 = 2 + 3C_{eq}$

$C_{eq}^2 - C_{eq} - 2 = 0$

Factoring the quadratic equation:
$(C_{eq} - 2)(C_{eq} + 1) = 0$

Since capacitance must be positive, we discard $C_{eq} = -1\ \mu\text{F}$.

Therefore, $C_{eq} = 2\ \mu\text{F}$.

Answer: $2\ \mu\text{F}$