Show that the following is indeed a mass function for R.V. $Y$ which can take values $2^n$ and $-2^n$ with probability $\frac{1}{2^{n+2}}$ So I think I have the pieces, just having trouble putting the puzzle together.
$P(Y=2^{n})=P(Y=-2^{n}) = \frac{1}{2^{n+2}}$
\begin{align*}
\sum_{n=0}^{\infty} p_y(y) 
&=\sum_{n=0}^{\infty} \frac{1}{2^{n+2}}\\
&=\frac{1}{4} \sum_{n=0}^{\infty} \frac{1}{2^n}\\
&=1/2
\end{align*}
Is this on the right track?  Is it $\sum_{n=0}^{\infty} p_y(y) =2*\sum_{n=0}^{\infty} \frac{1}{2^{n+2}} =\frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2^n}=1$ instead.
I also need to show that the expectation does not exist.  So I have
So far I have that $E(Y) = \sum_i y_ip_y(y_i)$, thus 
\begin{align*}
E(Y)= \sum_{n=0}^{\infty} 2^n \frac{1}{2^{n+2}} = \infty\\
E(Y)= \sum_{n=0}^{\infty} -2^n \frac{1}{2^{n+2}} = -\infty
\end{align*}
and thus the expectation does not exist
As always, thanks for any help.

Overall you are doing the problem correctly. The probability mass function has been given to you explicitly. Perhaps you were asked to show that it is indeed a pmf, though that is not clear from the question.  
If that was asked for, you need to sum the probabilities ("weights") and show that the sum is $1$. The sum of the weights at numbers of the form $2^n$ is $\sum_{n=0}^\infty \frac{1}{2^{n+2}}$, which you showed is $\frac{1}{2}$. The sum of the weights at numbers of the form $-2^n$ is also $\sum_{n=0}^\infty  \frac{1}{2^{n+2}}$, another $\frac{1}{2}$, for a total of $1$. So we were indeed given a pmf.
For the expectation, we need to find 
$$\sum_{n=0}^\infty 2^n \cdot\frac{1}{2^{n+2}} + \sum_{n=0}^\infty (-2^n) \cdot\frac{1}{2^{n+2}}.$$
 Both of the above sums diverge, since $2^n\cdot\frac{1}{2^{n+2}}=\frac{1}{4}$.
Thus the expectation does not exist.
This is an interesting example of a pmf which is symmetrical about $x=0$, but such that the mean is not $0$. In all symmetrical situations, if the mean exists, then it is at the centre of symmetry.    
