asymptotic distribution for MLE - Borel distribution Suppose we have sample $y_1,y_2,...y_n$ from a borel distribution
$$P(Y=y;\alpha)= \dfrac{1}{y!}(\alpha y)^{y-1}e^{-\alpha y} , y=1,2..$$ 
The MLE of $\alpha$ is $\hat{\alpha} = 1-\dfrac{1}{\bar{y}} $
By using that $\sqrt{n}\Big(\bar{y}  - \dfrac{1}{1-\alpha} \Big) \rightarrow N \Big(0,\dfrac{\alpha}{(1-\alpha)^3} \Big)$ 
How can I find the asymptotic distribution for $\hat{\alpha}$ (MLE)?
added extra text
My thought is that I can make use of the following: Set $\bar{y}(\alpha)= \dfrac{1}{1-\alpha}$. Then we have
$$\sqrt{n}\Big( \bar{y}({\hat{\alpha}}) - \bar{y}(\alpha)\Big) \rightarrow N(0,I(\bar{y})^{-1}). $$
$I(\alpha) = (\dfrac{d \bar{y} }{d\alpha})^2 \cdot I(\bar{y}(\alpha)) = \dfrac{1}{(1-\alpha)^4}  \cdot \dfrac{(1-\alpha)^3}{\alpha} = \dfrac{1}{(1-\alpha)\alpha} $  (by reparametrization lemma).
Finally we obtain:
$$\sqrt{n}\Big( \hat{\alpha} - \alpha\Big) \rightarrow N(0,I(\alpha)^{-1}) = N(0,(1-\alpha)\alpha) $$
but that is exactly the distribution you got :) @Ben
 A: From the form of your estimator, its distribution function can be written as:
$$F_\hat{\alpha}(a) = \mathbb{P}(\hat{\alpha} \leqslant a) = \mathbb{P} \Big( 1 - \frac{1}{\bar{Y}}  \leqslant a \Big) = \mathbb{P} \Big( \bar{Y} \leqslant \frac{1}{1 - a} \Big) = F_\bar{Y}\Big( \frac{1}{1 - a} \Big).$$
Let $\Phi$ denote the standard normal distribution function and $\phi$ denote the standard normal density function.  Using the asserted limiting result (corrected in the comment by dietervdf) you then have the asymptotic result:
$$\begin{equation} \begin{aligned}
F_\hat{\alpha}(a) 
&= \mathbb{P} \Big( \bar{Y} \leqslant \frac{1}{1 - a} \Big) \\[6pt]
&= \mathbb{P} \Big( \bar{Y} - \frac{1}{1 - \alpha} \leqslant \frac{1}{1 - a} - \frac{1}{1 - \alpha} \Big) \\[6pt]
&\approx \Phi \Bigg( \sqrt{n} \cdot \frac{\tfrac{1}{1 - a} - \tfrac{1}{1 - \alpha}}{\sqrt{\tfrac{\alpha}{(1-\alpha)^3}}} \Bigg) \\[6pt]
&= \Phi \Bigg( \sqrt{n \cdot \frac{1-\alpha}{\alpha}} \cdot \Big( \frac{1-\alpha}{1 - a} - 1 \Big) \Bigg). \\[6pt]
\end{aligned} \end{equation}$$
Differentiating and applying the chain rule yields the asymptotic density function:
$$\begin{equation} \begin{aligned}
f_\hat{\alpha}(a) 
&= \frac{dF_\hat{\alpha}}{da} (a)\\[6pt]
&\approx \sqrt{n \cdot \frac{1-\alpha}{\alpha}} \cdot \frac{1-\alpha}{(1 - a)^2} \cdot \phi \Bigg( \sqrt{n \cdot \frac{1-\alpha}{\alpha}} \cdot \Big( \frac{1-\alpha}{1 - a} - 1 \Big) \Bigg) \\[6pt]
&= \sqrt{\frac{n}{2 \pi} \cdot \frac{1-\alpha}{\alpha}} \cdot \frac{1-\alpha}{(1 - a)^2} \cdot \exp \Bigg( - \frac{n}{2} \cdot \frac{1-\alpha}{\alpha} \cdot \Big( \frac{1-\alpha}{1 - a} - 1 \Big)^2 \Bigg). \\[6pt]
\end{aligned} \end{equation}$$
