Find asymptotic variance MLE heavy tailed distribution $$\mathbf{X} = \{X_1,X_2,\dots,X_n\}$$ sequence of i.i.d. RV's. Let the distribution of the RV's be defined by
$$f(x|\theta)=\frac{\theta}{x^{\theta+1}}, \quad x>1, \quad \theta>1$$
I am interested in the asymptotic variance of the MLE for $\theta$. That is;
$$\lim_{n\rightarrow \infty}\mathbb{V}ar_{\theta^*}(\hat{\theta}_n) = \lim_{n\rightarrow \infty}\mathbb{V}ar_{\theta^*}\left( \frac{n}{\sum_{i=1}^n X_i}\right)$$ 
What I thought is to use the delta-method (1-st order) approximation:
$$\mathbb{V}ar_{\theta^*}(\hat{\theta}_n(\mathbf{X})) \approx 
\left(\hat{\theta}_n\left(\mathbb{E}_{\theta^*}[\mathbf{X}]\right) \right)^2\mathbb{V}ar_{\theta^*}(\mathbf{X})$$
In the above, subscript $\theta^*$ means with respect to the density defined by the true parameter value $\theta^*$ (assuming convergence in probability). Also, the expected value of one RV equals:
$$\mathbb{E}_{\theta^*}[X]=\frac{\theta}{1+\theta}$$
If I'm correct.
Would the calculation of the asymptotic variance be possible using the delta-method approximation? I could not get far at all.
 A: Your MLE for $\theta$ does not seem correct.  If $\boldsymbol x = (x_1, \ldots, x_n)$ is an iid sample drawn from a $X \sim \operatorname{Pareto}(1,\theta)$ distribution with density $$f_X(x) = \theta x^{-\theta-1} \mathbb{1}(x > 1)$$ then the likelihood function is $$\mathcal L (\theta \mid \boldsymbol x) = \theta^n \left(\prod_{i=1}^n x_i \right)^{-\theta-1} \mathbb{1}(x_{(1)} > 1).$$  The log-likelihood is then $$\ell(\theta \mid \boldsymbol x) = n \log \theta - (\theta+1) \sum_{i=1}^n \log x_i + \log \mathbb{1}(x_{(1)} > 1),$$ which suggests letting $\overline{ \log x} = \frac{1}{n} \sum_{i=1}^n \log x_i$.  Solving for the critical points gives $$0 = \frac{\partial \ell}{\partial \theta} = \frac{n}{\theta} - n \, \overline{\log x},$$ hence $$\hat \theta = \left( \overline{\log x} \right)^{-1} = \frac{n}{\sum_{i=1}^n \log x_i }.$$  This is plausible since the Pareto distribution has greater density in the tail when $\theta$ is small (this is obvious from inspecting the density); thus if the observations are large, the estimate of $\theta$ is small, but conversely, if the observations are close to $1$, the sum of their logarithms will be close to zero, yielding a large estimate of $\theta$ as this means the tail is relatively thin.  Your estimator clearly cannot work because it cannot give $\hat \theta > 1$.

Regarding the actual question of the asymptotic variance of this estimator, we first set out to find the distribution of the random variable $Y = \log X$.  Note this is a monotone transformation with the choice $g(x) = \log x$, hence we simply have $$f_Y(y) = f_X(g^{-1}(y)) \left| \frac{dg^{-1}}{dy} \right| = \theta \left( e^y \right)^{-\theta-1} e^y = \theta e^{-\theta y}, \quad y > 0.$$  We immediately recognize this as an exponential distribution with rate parameter $\theta$.  Therefore, the random variable $$S = \sum_{i=1}^n \log X_i \sim \operatorname{Gamma}(n, \theta),$$ being the sum of $n$ iid exponential distributions, where the parametrization used throughout is a rate:  $$f_S(s) = \frac{\theta^n s^{n-1} e^{-\theta s}}{\Gamma(n)}, \quad s > 0.$$  It easily follows that the distribution of the MLE estimator is $$\hat \theta = \frac{n}{S} \sim \operatorname{InvGamma}(n,n\theta),$$ with density $$f_{\hat\theta}(t) = \frac{(n \theta)^n e^{-n \theta/t}}{t^{n+1} \Gamma(n)}, \quad t > 0.$$  It is now simple to verify that the exact variance of the estimator is $$\operatorname{Var}[\hat\theta] = \frac{(n \theta)^2}{(n-1)^2 (n-2)}, \quad n > 2,$$ which asymptotically tends to $0$ as $n \to \infty$.
Moreover, we can use the above result to calculate the bias; clearly, for small sample sizes, $\hat \theta$ is biased, but for large sample sizes, this estimator has remarkably good properties as it is asymptotically unbiased, and its variance is $\mathcal O(n^{-1})$.
