How will $0<\theta <1$ influence this solution Given are $n$ samples $x_1,\ldots,x_n$ from the following distribution:
$$f(x;\theta)=\theta(1-\theta)^{x-1}$$
where $0<\theta<\frac{1}{2}$ and $x\in\{1,2,3\ldots\}$
Find the MLE estimator $\hat {\theta}_{MLE}$
I have found that $\hat {\theta}_{MLE}=\frac{1}{\left(1+\frac{1}{n}\sum_{k=1}^n(x_k-1)\right)}$ by the work shown in the end of this post.
How do I use the fact that $0<\theta<\frac{1}{2}$?
Please correct me if I made an error on the MLE calculation.
Finding the MLE:
$$\hat {\theta}_{MLE}=\arg{\max_{N}{(l(\theta))}}=\arg{\max_{N}{(\log(L(\theta)))}}=\arg{\max_{N}{(\log(\prod_{i=1}^nf(x_i|\theta)))}}$$
$$\hat {\theta}_{MLE}=\arg{\max_{N}{\left(\log(\prod_{i=1}^nf(x_i|\theta))\right)}}=\arg{\max_{N}{\left(\sum_{i=1}^n\log(f(x_i|\theta))\right)}}$$$$\hat {\theta}_{MLE}=\arg{\max_{N}{\left(\sum_{i=1}^n\log(\theta(1-\theta)^{x_i-1})\right)}}=\arg{\max_{N}{\left(n\log(\theta)+\sum_{i=1}^n\log(1-\theta)^{x_i-1}\right)}}$$$$\hat {\theta}_{MLE}=\arg{\max_{N}{\left(n\log(\theta)+\log(1-\theta)\sum_{i=1}^n(x_i-1)\right)}}$$
$$\frac{\mathrm d}{\mathrm d \theta} \left( n\log(\theta)+\log(1-\theta)\sum_{i=1}^n(x_i-1) \right)=\frac{n}{\theta}-\frac{\sum_{i=1}^n(x_i-1)}{1-\theta}=0$$$$(1-\theta)n=\theta \sum_{i=1}^n(x_i-1)$$
$$n=\theta \left(n+\sum_{i=1}^n(x_i-1)\right)$$
$$1=\theta \left(1+\frac{1}{n}\sum_{i=1}^n(x_i-1)\right)$$
$$\hat{\theta}_{MLE}=\frac{1}{1+\frac{1}{n}\sum_{i=1}^n(x_i-1)}$$
 A: So this answer is adding to what I commented. 
For  general version of the maximum likelihood estimator you need a little more than what it is written. If the parameter $\theta$ is known to take values in a set $ \Theta$ then
$$ \hat{\theta}_{MLE} = \arg \max_{ \theta \in \Theta} (\hat{l}(\theta)) $$
Note the difference of notation between $ l(\theta) $ and $ \hat{l} (\theta) $. The first one denotes the ''real likelihood'' while the second one considers using the sample. 
Now let's make sense of this idea: Suppose for example that you have want to estimate a parameter $ \theta$ which for some reason (probably nature of the experiment, some physical law, anything) is known to be in $ (0,1)$  . Then, imagine that you use the technique of derivation, and you get a formula for the estimator. But when you plug the value of the data into this formula it happens that your estimator is $1000$. Would that estimator make sense?
Of course not, so you have to restrict to maximize likelihood to the values for which it actually make sense. 
In your example, you need to verify that the value of your formula satisfies the restriction. But what if it does not? How would you know how  to compute the MLE?. It turns out that in many useful cases direct differentiation is not the most helpful example, in those cases you probably need to analyze the maximum with other techniques you should know from calculus. 
Note that 'the mistake' (it's not totally wrong) in your work is not considering the indicator functions. 
$$ \arg \max_{ \theta \in \Theta} \hat{l}(\theta) = \arg \max \prod_{i=1}^n f(x_i | \theta) \cdot 1_{\Theta}(\theta,x_i) $$
IN your particular example, as $x_i$ come from identical distributions we can simplify: 
$$ \arg \max_{ \theta \in \Theta} \hat{l}(\theta) =\arg \max \prod_{i=1}^n f(x_i | \theta) \cdot 1_{(0,1)}(\theta) =  1_{(0,1)}(\theta)\arg \max \prod_{i=1}^n f(x_i | \theta) $$
So I hope this explains what you needed. 
