Presumably the sum $S_m$ can be given in a more compact way as 
$S_m = 4\sum_{k=1}^{m-1} \frac{3^k} {2^{2^k}}.$ The limit should be a transcendental number, since the sum is extremely lacunary, so it seems unlikely that a closed for exists (since most closed form transcendental numbers are "named" constants").