Computing the Variance of an MLE 
Suppose we have i.i.d. $n$ observations $(X_1,X_2,...X_n)$ from a
  population with density $$f_\theta(x)=\begin{cases}\theta x^{\theta-1}&\text{ if }0\leq x\leq 1\\0&\text{otherwise.}\end{cases}$$ Note that $\theta>0$. If $T_n$ denotes the
  maximum likelihood estimator of $\theta$ given that $n$ is the sample
  size, then show that $$\text{var}(T_n)\stackrel{n\to\infty}{\longrightarrow} 0$$

This is the problem I am trying to solve: I computed $$T_n=\dfrac{-n}{\sum_{i=1}^{n}\ln(X_i)}$$ But I am stuck in finding $\text{var}(T_n)$ and showing that $\text{var}(T_n)\to0$ as $n\to\infty$.
 A: Combining @Xi'an comments and @wolfies' answer, we have that
$$1/T_n = \frac 1n \sum_{i=1}^n (-\ln X_i)$$
But $-\ln X_i \sim {\rm Exp}(1/\theta)$ (where $1/\theta$ is the scale parameter), which essentially is a Gamma distribution with shape parameter $1$, and so by the summation properties of the Gamma distribution,
$$\sum_{i=1}^n (-\ln X_i) \sim {\rm Gamma}(n, 1/\theta)$$.
Using the scaling properties of the Gamma distribution we obtain
$$1/T_n \sim {\rm Gamma}(n, 1/(n\theta)) $$
Then the inverse of $1/T_n$ (i.e. $T_n$) follows an Inverse Gamma distribution, with same shape paramater and reciprocal scale parameter
$$T_n \sim {\rm InvGamma}(n, n\theta)$$
that has variance
$$ {\rm Var} (T_n) = \frac {n^2\theta^2}{(n-1)^2(n-2)}$$
as the Mathematica software gave.
The leading term in the numerator is $n^2$ while the leading term in the denominator is $n^3$ so the limit of the variance of $T_n$ with respect to $n$ goes to zero.
A: Did you try to compute the Fisher information? Get the second derivative of the Log Likelihood, multiply by -1, and take the reciprocal. Evaluate at the MLE . Asymptotically, this approximates the variance of the MLE. 
A: The Problem
Let $(X_1, \dots, X_n)$ denote a random sample of size $n$ drawn on
$$X \sim \text{PowerFunction}(\theta,1)  \quad \text{ with pdf} \quad f(x) = \theta x^{\theta-1} \text{ where } \quad 0<x<1.$$
Let $Z  = -\sum_{i=1}^{n}\ln(X_i). \quad$ Find: $\quad Var\big[\dfrac{n}{Z}\big]$
Solution
Let $Y = -ln(X)$. Then, by any of the usual methods of transformation, $Y \sim \text{Exponential}(\theta)$ with pdf $\theta e^{-\theta y}$. Next, the sum of $n$ iid Exponentials with rate $\theta$ is well-known to be $\text{Gamma}(n, 1/\theta)$ so $Z \sim \text{Gamma}(n, 1/\theta)$ with pdf, say, $g(z)$:

We seek $Var(\frac{n}{Z})$:

where I am using the Var function from the mathStatica package for Mathematica to automate the calculation. 
The limit of the latter tends to 0, as $n \rightarrow   \infty$, as required.
