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Both $A$ and $B$ represent infinite geometric progression sum .

$A$ has a first term $a_1=1$ and a common ratio $r_1= \dfrac12$ , by applying the formula of an infinite geometric sum, we have $A = \dfrac {a_1}{1-r_1} = \dfrac1{1 - \frac12} = 2$ .

$B$ has a first term $a_2=1$ and a common ratio $r_2= -\dfrac12$ , by applying the formula of an infinite geometric sum, we have $B = \dfrac {a_2}{1-r_2} = \dfrac1{1 + \frac12} = \dfrac1{\frac32} = \dfrac23$ .

Hence, $3B = 2 = A$ .

Notice that both $A$ and $B$ can be evaluated because both their common ratio are in the interval $-1 < r < 1$ .

Relevant wiki: Geometric Progression Sum

$\begin{aligned} A & = 1 + \frac 12 + \frac 14 + \frac 18 + \frac 1{16} + \frac 1{32} + ... \\ & = \sum_{n=0}^\infty \left( \frac 12\right)^n & \small \color{#3D99F6}{\text{This is a sum of infinite GP}} \\ & = \frac 1{1-\frac 12} = 2 \end{aligned}$

$\begin{aligned} B & = 1 - \frac 12 + \frac 14 - \frac 18 + \frac 1{16} - \frac 1{32} + ... \\ & = 1 + \frac 14 + \frac 1{16} + ... - \left(\frac 12 + \frac 18 + \frac 1{32} + ... \right) \\ & = 1 + \frac 14 + \frac 1{16} + ... - \frac 12 \left(1 + \frac 14 + \frac 1{16} + ... \right) \\ & = \frac 12 \left(1 + \frac 14 + \frac 1{16} + ... \right) \\ & = \frac 12 \sum_{n=0}^\infty \left( \frac 14\right)^n \\ & = \frac 12 \cdot \frac 1{1-\frac 14} = \frac 23 \end{aligned}$

$\implies \boxed{A=3B}$