The envelope distribution where switching always wins

The naive "always switch" argument for the two-envelope problem is usually wrong because no proper prior makes the switch gain positive at every observed value. But suppose the smaller amount is $A = 2^n$ (so the pair is $\{2^n, 2^{n+1}\}$) with probability

$P(A = 2^n) = \tfrac13 \left(\tfrac23\right)^n, \qquad n = 0, 1, 2, \dots$

You open your envelope and see $X = 2^m$.

Show that switching has positive expected gain for every value $X$ you could observe. Then explain why this does not actually let you "switch forever" for a contradiction.