\displaystyle  \sum_{n=1}^{\infty} \frac{1}{2^{2\log_2 n + C}} = \frac{1}{2^C}\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6\cdot 2^C} < \frac{1}{2^{C-1}}. 