real analysis - series is divergent

I want to prove that $\sum_{k=1}^{\infty}(\frac{1}{k}-\frac{1}{2^k})$ diverges? I want to do this with the comparison test, but I don't find a divergent series.

An other point I want to ask, if you can do this: $\sum_{k=1}^{\infty}(\frac{1}{k}-\frac{1}{2^k})=\sum_{k=1}^{\infty}\frac{1}{k}-\sum_{k=1}^{\infty}\frac{1}{2^k}$? One of this series of the right hand side is convergent, therefore you don't have the undefined case $\infty -\infty$.