I am trying to calculate the mean and variance of the negative binomial variable using the parameterization $NB(r, \mu)$, where $\mu$ is the mean of the variable. That is, I am trying to proof that $$\mu = E(X), $$ where $X \sim NB(r, \mu)$. I tried the following
\begin{align} E(X) &= \sum_{x = 0}^\infty x \binom{x + r - 1}{x} \left( \frac{\mu}{r + \mu} \right)^x \left( \frac{r}{r + \mu} \right)^r\\ &= \sum_{x = 1}^\infty \frac{(x + r - 1)!}{(x - 1)! (r - 1)!} \left( \frac{\mu}{r + \mu} \right)^x \left( \frac{r}{r + \mu} \right)^r\\ &= r \left( \frac{\mu}{r + \mu} \right) \left( \frac{r + \mu}{r} \right) \sum_{x = 1}^\infty \frac{(x + r - 1)!}{(x - 1)!r!} \left( \frac{\mu}{r + \mu} \right)^{x - 1} \left( \frac{r}{r + \mu} \right)^{r + 1}\\ &= \mu \sum_{x = 1}^\infty \frac{(x + r - 1)!}{(x - 1)!r!} \left( \frac{\mu}{r + \mu} \right)^{x - 1} \left( \frac{r}{r + \mu} \right)^{r + 1}\\ &= \mu \sum_{y = 0}^\infty \frac{(y + r)!}{y!r!} \left( \frac{\mu}{r + \mu} \right)^y \left( \frac{r}{r + \mu} \right)^{r + 1}, \hspace{20pt} \mbox{where } y = x - 1\\ &= \mu \sum_{y = 0}^\infty \frac{(y + s - 1)!}{y! (s - 1)!} \left( \frac{\mu}{s - 1 + \mu} \right)^y \left( \frac{s - 1}{s - 1 + \mu} \right)^s, \hspace{20pt} \mbox{where } s = r + 1\\ &= \mu \sum_{y = 0}^\infty \binom{y + s - 1}{y} \left( \frac{\mu}{s - 1 + \mu} \right)^y \left( \frac{s - 1}{s - 1 + \mu} \right)^s \end{align}
Then I am stuck. I know that if I use the parameterization $NB(r, p)$, my life will be much better as the I have to only deal with $p$ instead of $\dfrac{\mu}{r + \mu}$, I do not have to worry how should I modify the fraction when I let $s = r + 1$, $y = x - 1$.