Maximum Likelihood Estimator for Negative Binomial Distribution The question is the following:

A random sample of n values is collected from a negative binomial distribution with parameter k = 3.
  
  
*
  
*Find the maximum likelihood estimator of the parameter π.
  
*Find an asymptotic formula for the standard error of this estimator.
  
*Explain why the negative binomial distribution will be approximately normal if the parameter k is large enough. What are the parameters of this normal approximation?
  

My working has been the following:
 1. I feel like this is what is wanted but I'm not sure if I'm accurate here or if I can possibly take this further given the information provided?
$$p(x) = {x-1 \choose k-1}\pi^k(1-\pi)^{x-k}\\
L(\pi) = \Pi_i^n p(x_n|\pi)\\
\ell(\pi) = \Sigma_i^n\ln(p(x_n|\pi))\\
\ell`(\pi) = \Sigma_i^n\dfrac{k}{\pi}-\dfrac{(x-k)}{(1-\pi)}$$


*I think the following is what is asked for. For the final part I feel like I need to replace $\hat{\pi}$ with $\dfrac{k}{x}$
$$\ell``(\hat{\pi}) = -\dfrac{k}{\hat{\pi}^2} + \dfrac{x}{(1-\hat{\pi})^2}\\
se(\hat{\pi}) = \sqrt{-\dfrac{1}{\ell``(\hat{\pi})}}\\
se(\hat{\pi}) = \sqrt{\dfrac{\hat{\pi}^2}{k} - \dfrac{(1-\hat{\pi})^2}{x}}\\$$

*I am not really sure how to prove this one and am still researching it. Any hints or useful links would be greatly appreciated. I feel like it is related either to the fact that a negative binomial distribution can be seen as a collection of geometric distributions or the inverse of a binomial distribution but not sure how to approach it.
Any help at all would be greatly appreciated
 A: 1.
$p(x) = {x_i-1 \choose k-1}\pi^k(1-\pi)^{x_i-k}$
$L(\pi;x_i) = \prod_{i=1}^{n}{x_i-1 \choose k-1}\pi^k(1-\pi)^{x_i-k}\\$ 
$
\ell(\pi;x_i) = \sum_{i=1}^{n}[log{x_i-1 \choose k-1}+klog(\pi)+(x_i-k)log(1-\pi)]\\
\frac{d\ell(\pi;x_i)}{d\pi} = \sum_{i=1}^{n}[\dfrac{k}{\pi}-\dfrac{(x_i-k)}{(1-\pi)}]$
Set this to zero,
$\frac{nk}{\pi}=\frac{\sum_{i=1}^nx_i-nk}{1-\pi}$
$\therefore$ $\hat\pi=\frac{nk}{\sum_{i=1}^nx}$

2. 


For second part you need to use the theorem that $\sqrt{n}(\hat\theta-\theta) \overset{D}{\rightarrow}N(0,\frac{1}{I(\theta)})$, $I(\theta)$ is the fisher information here. Therefore,the standard deviation of the $\hat\theta$ will be $[nI(\theta)]^{-1/2}$. Or  you call it as standard error since you use CLT here.
So we need to calculate the Fisher information for the negative binomial distribution.
$\frac{\partial^2 \log(P(x;\pi))}{\partial\pi^2}=-\frac{k}{\pi^2}-\frac{x-k}{(1-\pi)^2}$
$I(\theta)=-E(-\frac{k}{\pi^2}-\frac{x-k}{(1-\pi)^2})=\frac{k}{\pi^2}+\frac{k(1-\pi)}{(1-\pi)^2\pi}$
Note: $E(x) =\frac{k}{\pi}$ for the negative binomial pmf 
Therefore, the standard error for  $\hat \pi$ is $[n(\frac{k}{\pi^2}+\frac{k(1-\pi)}{(1-\pi)^2\pi})]^{-1/2}$
Simplify we get we get $se(\pi)=\sqrt{\dfrac{\pi^2(\pi-1)}{kn}}$

3. 


The geometric distribution is a special case of negative binomial distribution when k = 1. Note $\pi(1-\pi)^{x-1}$ is a geometric distribution
Therefore, negative binomial variable can be written as a sum of k independent, identically distributed (geometric) random variables.
So by CLT negative binomial distribution will be approximately normal if the parameter k is large enough
