Example 7: Suppose that we flip a coin until a tail first appears and if the number of tosses equals $k$, then we are paid $2^k$ dollars. What is $E[U(2^k)]$, where $ U(p)= \frac{p^{1-\gamma} -1}{1- \gamma}$.
I believe $E[U(2^k)] = \sum_{k=1}^{\infty} \frac{1}{2} ^k * ( \frac{2^{k^{1-\gamma}} -1}{1- \gamma})$
But, is there a way to simplify this? Specifically, is there a way to get rid of $K$ so that we are only left with one variable, $\gamma$?
\begin{align} \frac{1}{1-\gamma}\sum_{k=1}^\infty \left( \frac{2^{k(1-\gamma)}}{2^k} - \frac1{2^k} \right) &= \frac{1}{1-\gamma}\sum_{k=1}^\infty \left( 2^{-k\gamma} - \frac1{2^k} \right) \end{align}
Now, on the right, it it just two geometric series. Hopefully you can take it from here.