CRLB/UMVUE estimation of $\theta$ 
We have a random sample $X_1,X_2,\ldots,X_n$ from a probabilitiy distribution with density $f(x;\theta) = \theta x^{-\theta-1} $ given that $x > 1$, and $0$ else. where $\theta >1 $ is an unknown estimator
  
  
*
  
*derive the CRLB for unbiased estimators of $\theta$.
  
*Is there an unbiased estimator of $\theta$ that reaches the CRLB?
  
*show that $S = \sum^n_{i=1} \ln(X_i)$ is a complete sufficient statistic and use this result to derive thje UMVUE for $\frac{1}{\theta}$

for 1. i did the following:
$$
\begin{align}
f_\bar{x}(x_1,...,x_n) &= \prod_{i=1}^n \theta x_i^{-(\theta+1)} \\ 
l(\theta) &= n\ln(\theta) - (\theta+1) \sum_{i=1}^n \ln(x_i) \\
\frac{\delta}{\delta \theta} l(\theta) &= -\frac{\theta+1}{\sum_{i=1}^n x_i}
\end{align}
$$
however setting this to 0 really doesnt help me. I have been told that looking at the $l(\theta)$ function and seeing how it behaves should help, but I really am not getting any wiser.
for 2. I would just fill in the CRLB equation, should i use $\tau(\theta) = \hat{\theta}$ ?
for 3. what is the difference between a complete sufficient statistic and a sufficient statistic?
 A: This is not the complete solution, but maybe helpful on your way.
What you have is the Pareto distribution with the scale parameter $x_m=1$.

First of all you have an error while computing the derivative of the log-likelihood function.
$$l(\theta) = n\ln(\theta) - (\theta+1) \sum_{i=1}^n \ln(x_i)$$
Taking the derivative with the respect to $\theta$ one can get:
$$\frac{\partial l(\theta) }{\partial \theta}=\frac{n}{\theta}-\sum_{i=1}^n \ln(x_i)$$
Second, you do not need to set it to zero (unless you are after the ML estimator). Moreover to compute the Cramer-Rao bound you need to take the second derivative:
$$\frac{\partial^2 l(\theta) }{\partial \theta^2}=-\frac{n}{\theta^2}$$
Then taking the expectation:
$$\mathrm{E}\left\{ \left(\frac{\partial l(\theta) }{\partial \theta}\right)^2\right\}=-\mathrm{E}\left\{ \frac{\partial^2 l(\theta) }{\partial \theta^2}\right\}=\frac{n}{\theta^2}\int_1^\infty \theta x^{-\theta-1} \mathrm dx=\frac{n}{\theta^2}$$
Here as Lost1 indicated we are asked to find the bound for the parameters' estimate not its' function, so: $$\mathrm var(\theta)\geq\frac{\theta^2}{n}$$


By the way if you need the ML estimator then you can set the first derivative to zero and get:
$$\hat{\theta}_{ML}:\left.\frac{\partial l(\theta) }{\partial \theta}\right\vert_{\theta=\hat{\theta}_{ML}}=0 \quad \Rightarrow \quad\hat{\theta}_{ML}=\frac{n}{\sum_{i=1}^n \ln(x_i)}$$
