unbiased estimator for the negative binomial distribution Let be $P(X=x)= \binom{x-1}{r-1} \theta^r (1-\theta)^{x-r}$
Show that $\frac{r-1}{x-1}$ is an unbiased estimator for 
My attempt;
E$[\frac{r-1}{x-1}]$ $=$ $\sum_{x=0}^{\infty}\frac{r-1}{x-1}\binom{x-1}{r-1} \theta^r (1-\theta)^{x-r} = \sum_{x=0}^{\infty} \binom{x-2}{r-2} \theta^{r-1} (1-\theta)^{x-r+1}$
$\sum_{x=0}^{\infty} \binom{x-1-1}{r-1-1} \theta^{r-1} (1-\theta)^{x-r+1}$
Such that $E(x)$ from a $BinNeg (r,p)$ is $E(x)= \frac{r(1-p)}{p}$ then;
E$[\frac{r-1}{x-1}]$ must be $\frac{(r-1)(1-p)}{p}$
So that means that $\frac{r-1}{x-1}$ is not an unbiased estimator for 
Is that right?
I'm not sure about the value of r

There are some errors in your calculation.  Note that the lower index of summation should be $r$, not $0$.  Then we have
$$\begin{align}
\frac{r-1}{x-1} \binom{x-1}{r-1} \theta^r (1-\theta)^{x-r} 
&= \frac{r-1}{x-1} \frac{(x-1)!}{(r-1)! (x-r)!} \theta^r (1-\theta)^{x-r} \\
&= \theta \frac{(x-2)!}{(r-2)! ((x-2)-(r-2))!} \theta^{r-1} (1-\theta)^{(x-1)-(r-1)} \\
&= \theta \binom{(x-1)-1}{(r-1)-1} \theta^{r-1} (1-\theta)^{(x-1)-(r-1)}.\end{align}$$
Note this last expression is $\theta$ times the PMF of a negative binomial distribution with parameters $r^* = r-1$ and $\theta$.  Hence
$$\operatorname{E}[X] = \theta \sum_{x=r}^\infty \binom{(x-1)-1}{(r-1)-1} \theta^{r-1} (1-\theta)^{(x-1)-(r-1)} = \theta \sum_{x=r^*}^\infty \binom{x-1}{r^*-1} \theta^{r^*} (1-\theta)^{x-r^*} = \theta.$$
