Is there any way to simplify this term: $$\sum_{v=1}^m v(1-\frac{1}{v})^u\binom{m}{v}p^v(1-p)^{m-v}$$ The term $(1-\frac{1}{v})^u$ is really annoying.
This expansion appear in a specific version of the urn and ball game. Suppose there are $u$ balls that will be randomly allocated to $V$ urns with $V$ being a random variable with binomial distribution $B(m,p)$.
Then the expected number of urns with at least one ball should be: $$ \sum_{v=0}^m v(1-(1-\frac{1}{v})^u) \binom{m}{v} p^v (1-p)^{m-v} $$
Break the expansion into two parts. The first part: $$ \begin{align} \sum_{v=0}^m v\binom{m}{v}p^v(1-p)^{m-v} &= mp \sum_{v=1}^m \binom{m-1}{v-1}p^{v-1}(1-p)^{m-v} \\ & = mp \end{align} $$ My question is how to deal with the second part.
