Strong convergence of probability measures implies absolute continuity? Suppose that $(\mu_n)_n$ is a sequence of probability measure for which $\lim_n\langle\phi,\mu_n\rangle=\langle\phi,\mu\rangle$ where $\mu$ is a probability measure and $\phi$ is any bounded, real valued, measurable function. I want to show that if $\sum_{n\in\mathbb{N}}2^{-n}\mu_n(A)=0$ then $\mu(A)=0$.

Unless I am missing something, this seems pretty straightforward.  Since all the $\mu_n$ are probability measures, in particular we have $\mu_n(B) \ge 0$ for every $B$.  So if $A$ is a measurable set such that $\sum 2^{-n} \mu_n(A) = 0$ then we must have $\mu_n(A) = 0$ for every $n$.  Now applying your convergence statement with the indicator function $\phi = 1_A$, we have
$$\mu(A) = \langle 1_A, \mu \rangle = \lim \langle 1_A, \mu_n \rangle = \lim \mu_n(A) = \lim 0 = 0.$$
