4.4.2 problem 63

$\text{E}\left(2^X\right)={\sum }_{k=0}^{\infty }{2}^k{(1-p)}^{k}p=p{\sum }_{k=0}^{\infty }{(2-2p)}^k=\frac{p}{2p-1}$ when $2-2p<1\Rightarrow p>\frac{1}{2}$.

$\text{E}\left(2^{-X}\right)={\sum }_{k=0}^{\infty }{2}^{-k}{(1-p)}^{k}p=p{\sum }_{k=0}^{\infty }{\left(\frac{1-p}{2}\right)}^k=\frac{p}{\frac{1+p}{2}}=\frac{2p}{1+p}$ when $\frac{1-p}{2}<1$ which is always true.