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.
 A: 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.    
