2. It looks like you already have it! You wrote down $\displaystyle \sum_{n=1}^\infty 2^n\frac{1}{2^n \log (2^n)}$. This is precisely $\displaystyle \sum_{n=1}^\infty \frac{1}{\log (2^n)}=\sum_{n=1}^\infty \frac{1}{n \log 2}=\displaystyle \frac{1}{\log 2}\sum_{n=1}^\infty \frac{1}{n}$, which diverges.