what's the distribution of the inverse of a random variable that follows a negative binomial distribution?

I was studying the method of moments estimation of parameters, and I encountered the following problem.

I have a geometric distribution as following:

$P(X=k) = p(1-p)^{k-1}$, and a sample size of n, all observations are independent

Hence, it can be quite easily shown that

$\hat p = \dfrac{1}{\overline{X}}$, where $\overline{X}$ is the sample mean, this is the MME estimation of p.

Now, if I want to find the approximate normal distribution for $\hat p$, I would have to find out the variance of $\hat p$

Since, $\hat p = \dfrac{n}{\sum\limits_{i=1}^n X_n}$, we see that the bottom follows a negative binomial distribution $NegativeBin(n, p)$.

Now I am stuck at this step, and I can't figure out how to calculate the expression for the PMF and the variance of $\hat p$.

Thank you very much!

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we see that the bottom follows a binomial distribution Bin(n,p)... Do we, really? Note that the $X_i$s are not 0-1 valued. – Did Feb 10 '13 at 21:02
Sorry, I meant negative binomial, I will edit the post – Enzo Feb 10 '13 at 21:15

Here is a fast Ansatz to find the approximate normal distribution for $\hat p_n$ when $n\to\infty$.
By the CLT, $\sum\limits_{i=1}^nX_i=nm+\sqrt{nv}Z_n$ where $m=\mathbb E(X_1)$, $v=\mathrm{var}(X_1)$ and $Z_n$ converges in distribution to a standard normal random variable. Thus, loosely speaking, $$\hat p_n=\frac1m\left(1+\frac{\sqrt{v}}{m\sqrt{n}}Z_n\right)^{-1}\approx \frac1m-\frac{\sqrt{v}}{m^2\sqrt{n}}Z_n.$$ In particular, one may hope that $$\mathbb E(\hat p_n)\approx\frac1m,\quad\mathrm{var}(\hat p_n)\approx\frac{v}{m^4n}.$$ This suggests that $$\hat p_n=\frac1m+\frac{\sqrt{v}}{m^2\sqrt{n}}V_n,$$ where $V_n$ converges in distribution to a standard normal random variable.