Poisson distribution and price reduction A store owner has certain existence of an item and he decides to use the following scheme to sell it:
The item has a price of \$100. The owner will reduce the price in half for every customer that buys the item on a given day. That way, the first customer will pay \$50, the next one will pay $25, and so on... 
Suppose that the number of customers who buy the item during a day follow a Poisson distribution, with $\mu = 2$ Find the expected price of the item at the end of a day.

I defined a "final price" function like this:
$$g(x) = \frac{100}{2^x}$$
where $x$ is the random variable of the Poisson distribution:
$$f(x)=\frac{\lambda^xe^{-x}}{x!}$$
Since $\mu = \lambda = 2$, I can rewrite $f(x)$ like this:
$$f(x) = \frac{2^x}{e^2x!}$$
And there's where I'm stuck. I see two possible roads:
I'll use capital $X$ for the random variable and lower-case $x$ for the argument to the probability mass function (thus if I write $\Pr(X=x)$ with both capital $X$ and lower-case $x$ you'll know what I mean).  You have
\begin{align}
& \operatorname{E}\left( \frac {100}{2^X}\right) = \sum_{x=0}^\infty \frac{100}{2^x} \Pr(X=x) = \sum_{x=0}^\infty \frac{100}{2^x}\cdot \frac{2^x}{x!e^2} \\[10pt]
= {} & \frac{100}{e^2}\sum_{x=0}^\infty \frac 1 {x!} = \frac{100}{e^2}\cdot e = \frac{100} e \approx 36.78794.
\end{align}
