The E[x] equation you come up with in this problem can be simplified (rid the summation term) by using a differentiated form of the commonly used geometric series equation: $ \sum_{n=0}^\infty r^n = 1/(1-r) $
Now take the derivative with respect to r and you get:
$ \sum_{n=0}^\infty nr^{n-1} = 1/(1-r)^2 $
You can use this equation to simplify your expected value.
