Answer
$\dfrac{\pi }{{12}}$
Work Step by Step
From Figure 1, we see that the diameters of the circles (starting from the right one) are
$\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\cdot\cdot\cdot$
Thus, the total area is
$S = \pi \left[ {{{\left( {\dfrac{1}{2}\cdot\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{2}\cdot\dfrac{1}{4}} \right)}^2} + {{\left( {\dfrac{1}{2}\cdot\dfrac{1}{8}} \right)}^2} + \cdot\cdot\cdot} \right]$
$S = \dfrac{\pi }{4}\left[ {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{4}} \right)}^2} + {{\left( {\dfrac{1}{8}} \right)}^2} + \cdot\cdot\cdot} \right]$
$S = \dfrac{\pi }{4}\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{{2^{2n}}}} = \dfrac{\pi }{4}\mathop \sum \limits_{n = 1}^\infty {\left( {\dfrac{1}{4}} \right)^n}$
Notice that this is a sum of geometric series with ratio $r = \dfrac{1}{4}$.
By equation (5) of Theorem 2 in Section 11.2:
$S = \dfrac{\pi }{4}\left( {\dfrac{{\dfrac{1}{4}}}{{1 - \dfrac{1}{4}}}} \right) = \dfrac{\pi }{{12}}$
Thus, the total area of the infinitely many circles is $\dfrac{\pi }{{12}}$.
