=\frac{1}{2}\sum_{m=1}^{\infty}\sum_{n=1}^\infty \frac{1}{a_ma_n}=\frac{1}{2}\sum_{m=1}^{\infty}\sum_{n=1}^\infty \frac{mn}{2^m2^n}=\frac{1}{2}\Big(\sum_{m=1}^{\infty}\frac{m}{2^n}\Big)^2=\frac{4}{2}=2.