We perform Bernoulli trials until $r$ failures ($P.(\text{fail}) = u$) occur. What is the probability that $k$ successes occur? \begin{align*} p(k) = {r + k - 1 \choose k} u^r (1 - u)^k \end{align*}
What is the expected number of successes before $r$ failures occur? \begin{align*} &\quad \mathbb{E}(p(k)) \\ &= \sum_{k = 0}^{\infty} k p(k) \\ &= \sum_{k = 0}^{\infty} k {r + k - 1 \choose k} u^r (1 - u)^k \\ &= \sum_{k = 0}^{\infty} k \frac{(r + k - 1)!}{k! (r + k - 1 - k)!} u^r (1 - u)^k \\ &= \sum_{k = 0}^{\infty} k \frac{(r + k - 1)!}{k! (r - 1)!} u^r (1 - u)^k \\ &\text{$0! = 1$, so when $k = 0$, the summand is $0$} \\ &= \sum_{k = 1}^{\infty} k \frac{(r + k - 1)!}{k! (r - 1)!} u^r (1 - u)^k \\ &= \sum_{k = 1}^{\infty} \frac{(r + k - 1)!}{(k - 1)! (r - 1)!} u^r (1 - u)^k \\ &= (1 - u)\sum_{k = 1}^{\infty} \frac{(r + k - 1)!}{(k - 1)! (r - 1)!} u^r (1 - u)^{(k - 1)} \\ &\text{let $j = k - 1$} \\ &= (1 - u) \sum_{j = 0}^{\infty} \frac{(r + j)!}{j! (r - 1)!} u^r (1 - u)^{j} \\ &= r(1 - u) \sum_{j = 0}^{\infty} \frac{(r + j)!}{j! r!} u^r (1 - u)^{j} \\ &= r(1 - u) \sum_{j = 0}^{\infty} {r + j \choose j} u^r (1 - u)^{j} \\ &= r(1 - u) \lim_{\alpha \rightarrow \infty} \sum_{j = 0}^{\alpha} {r + j \choose j} u^r (1 - u)^{j} \\ &\text{let $r + j = \alpha$} \\ &= r(1 - u) \lim_{\alpha \rightarrow \infty} \sum_{j = 0}^{\alpha} {\alpha \choose j} u^{\alpha - j} (1 - u)^{\alpha - r} \\ &= \frac{r(1 - u)}{(1 - u)^r} \lim_{\alpha \rightarrow \infty} \sum_{j = 0}^{\alpha} {\alpha \choose j} u^{\alpha - j} (1 - u)^{\alpha} \\ &= \frac{r}{(1 - u)^{r - 1}} \lim_{\alpha \rightarrow \infty} (u + 1 - u)^\alpha \\ &= \frac{r}{(1 - u)^{r - 1}} \lim_{\alpha \rightarrow \infty} 1^{\alpha} \\ &= \frac{r}{(1 - u)^{r - 1}} (1) \\ &= \frac{r}{(1 - u)^{r - 1}} \end{align*}
However, Wikipedia states that the expectation should be $(1 - u)r/u$. Where did I make an error in the above calculation?
