Series with powers of one half via a uniform expectation

Let $U$ be uniform on $(0,\tfrac12)$. By evaluating $\mathbb{E}\left[\frac{1}{1-U}\right]$ in two ways, find $\sum_{n\ge0}\frac{1}{2^n(n+1)}$. The result equals $\ln a$ for an integer $a$; give $a$.

$a=4$. Directly, $\mathbb{E}\tfrac{1}{1-U}=\int_0^{1/2}\tfrac{2}{1-u}du=2\ln2=\ln4$. Expanding as a geometric series, $\sum_k\mathbb{E}U^k$ with $\mathbb{E}U^k=\tfrac{1}{2^k(k+1)}$ reproduces the given series.