I aways seem to have a knack for finding a route that's a bit more complicated.
My thinking was the sides of the squares were $\displaystyle \frac{s}{\sqrt{2}}$, so their areas were 1/2, and the areas of pairs of shaded triangles were 1/4.
So I put down this series...
$\displaystyle \frac{1}{4}\sum(\frac{s}{(\sqrt{2})^{n-1}})^2$
Which I see now, can be simplified to $\displaystyle \sum\frac{s^2}{2^{n+1}}$, the same thing Spimon shows above.
-Scott