Let's cheat and use one of Euler's many results:
$\sum_{i=1}^{n} \frac{1}{i} = \ln n + \gamma + \frac{1}{2n} + O\left(\frac{1}{n^2}\right)$
Note that:
$A_n + \sum_{i=1}^{n} \frac{1}{i} = \sum_{i=1}^{2n} \frac{1}{i}$
Substituting Euler's result for both summations, we get:
$A_n + \ln n + \gamma + \frac{1}{2n} + O\left(\frac{1}{n^2}\right) = \ln 2n + \gamma + \frac{1}{4n} + O\left(\frac{1}{n^2}\right)$
Rearranging, and using $\ln 2n = \ln 2 + \ln n$, we get
$A_n = \ln 2 - \frac{1}{4n} + O\left(\frac{1}{n^2}\right)$
Thus the requested limit becomes
$\lim_{n \to \infty} n (\ln 2 - A_n) = \lim_{n \to \infty} n \left(\frac{1}{4n} - O\left(\frac{1}{n^2}\right)\right) = \frac{1}{4}$