(Feller Volume 1, Q.33, p.241) A sequence of Bernoulli trials is continued as long as necessary to obtain $r$ successes, where $r$ is a fixed integer. Let $X$ be the number of trials required. Find $E(r/X)$.
Solution: \begin{equation} E(X/r) = r \sum_{r=k}^\infty k^{-1} {k-1 \choose r-1} p^r q^{k-r}\\ = \sum_{k=1}^{r-1} (-1)^{k-1} \frac{r}{r-k} (p/q)^k + (-p/q)^r r \log p. \end{equation} To derive the last formula from the first, put $f(q) = r\sum k^{-1}{k-1 \choose r-1} q^k$. Using the identity ${-a \choose k} = (-1)^k{a+k-1 \choose k}$ for any $a>0$, we find that $f'(q) = nq^{r-1} (1-q)^{-r}$. The assertion now follows by repeated integrations by parts.
I think that there is some errata in the solution. The first display should be $$E(r/X) = r \sum_{k=r}^\infty k^{-1} {k-1 \choose r-1} p^r q^{k-r}. $$ I am confused about the next step. If I differentiate $f(q)$ with respect to $q$, I get $r \sum_{r=k}^\infty {k-1 \choose r-1} q^{k-1}$. I don't understand how the author gets $rq^{r-1}(1-q)^{-r}$. Also, how does this help to show the last formula from the second one.
I would appreciate if you give some help.