Virgil is right, it is
$ \sum_{i=1}^n\frac{n}{n - i + 1}\! $
because E[X] is 1/p
More over this can be simplified using the arithmetic series
$ \sum_{i=1}^n a_i = \frac{n(a_1+a_n)}{2}\! $
Virgil is right, it is
$ \sum_{i=1}^n\frac{n}{n - i + 1}\! $
because E[X] is 1/p
More over this can be simplified using the arithmetic series
$ \sum_{i=1}^n a_i = \frac{n(a_1+a_n)}{2}\! $