Let $y_1, \dots,y_n$ be i.i.d. random variables from $Y$ with density $$p(y;\theta)=\theta(1-y)^{\theta-1},\ \theta > 0,\ y \in(0,1)$$
and $T=-\ln(1-Y)\sim Exp(\theta)$ with $\theta$ rate parameter.
The log-likelihood is $l(\theta)=n \ln(\theta)+\sum\ln(1-y)$ and the MLE is $\hat \theta=-\frac{n}{\sum \ln(1-y)}$.
I need to prove the (weak) consistency of $\hat \theta$. A sufficient condition is
$$\begin{cases} \lim_{n \to \infty}\mathbb{E}(\hat \theta)=\theta \\ \lim_{n \to \infty}\mathbb{Var}(\hat \theta)=0 \end{cases}$$
Based on MLE invariance I could write $\hat \theta=\frac{n}{\sum t}$, so $$\lim_{n \to \infty}\mathbb{E}(\hat \theta)=\lim_{n \to \infty}\mathbb{E}\left(\frac{n}{\sum t}\right)=\lim_{n \to \infty}\left(n\frac{1}{n\frac{1}{\theta}}\right)=\theta$$
and $$\mathbb{Var}(\hat \theta)=\mathbb{Var}\left(\frac{n}{\sum t}\right)=n^2\mathbb{Var}\left(\frac{1}{\sum\mathbb{Var}(t)}\right)=n^2\frac{1}{\frac{n}{\theta^2}}=n\theta^2$$
What am I doing wrong?
Edit with solution
My solution: if $\sum t_i \sim \Gamma(n,\theta)$ then $\frac{1}{\sum t_i}\sim IG(n,\theta)$. Now, the variance of $IG$ should be $\mathbb{Var}(IG(n,\theta))=\frac{\theta^{2}}{(n-1)^2(n-2)}$ so $\lim_{n \to \infty}n^2\frac{\theta^{2}}{(n-1)^2(n-2)}=0$ and this prove the consistency of $\hat \theta$.
