We have a simple random sample of size n from a distribution with pmf () = $\theta{(1-\theta)}^{x-1}$ for = 1,2, …. Find the MLE[]
My try:
$ L\ =\ \theta{(1-\theta)}^{1-1}\times\theta{(1-\theta)}^{2-1}\times\ ...\ \theta{(1-\theta)}^{n-1}\ =\ \theta^n\prod_{i=1}^{n}{(1-\theta)}^{x_i-1} $
$ \log{L\left(\theta\right)}\ =\ \log{\left[\theta^n\prod_{i=1}^{n}\left(1-\theta\right)^{x_i-1}\right]}=\ n\log{\theta+\sum_{i=1}^{\infty}{\log{\left(1-\theta\right)}}^{x_i-1}=n\log{\theta+\log{\left(1-\theta\right)\sum_{i=1}^{\infty}{x_i-1}}}} $
$ \frac{d\ log\ \theta}{d\theta}\ =\ \frac{n}{\theta}\ +\ \frac{1}{\theta-1}\sum_{i=1}^{\infty}{x_i-1} $
I think something is incorrect here. How to simplify it further? Help please
Your steps are right so far. You have
$$n\log{\theta+\log{\left(1-\theta\right)\sum_{i=1}^{\color{red}n}{\color{red}(x_i-1\color{red})}}}=0$$
I've made some improvements (red). Writing for each summand a sigma sign.
$$n\log\theta+\log\left(1-\theta\right)\sum_{i=1}^{n} x_i-\log\left(1-\theta\right)\sum_{i=1}^{n}1$$
$$n\log\theta+\log\left(1-\theta\right)\sum_{i=1}^{n} x_i-\log\left(1-\theta\right)\cdot n$$
$$\frac{d \log L}{d \theta}=\frac{n}{\theta}-\frac{\sum\limits_{i=1}^{n} x_i}{1-\theta}+\frac{n}{1-\theta}=0$$
$$\frac{n}{\theta}=\frac{\sum\limits_{i=1}^{n} x_i}{1-\theta}-\frac{n}{1-\theta}$$
$$\frac{n}{\theta}=\frac{\sum\limits_{i=1}^{n} x_i-n}{1-\theta}$$
It is $\sum\limits_{i=1}^{n} x_i=n\cdot \overline x$. Thus we get
$$\frac{n}{\theta}=\frac{n\cdot \overline x-n}{1-\theta}$$
Dividing the equation by $n$
$$\frac{1}{\theta}=\frac{\overline x-1}{1-\theta}$$
$1.$ Multiplying the equation by $(1-\theta) \quad $ $2.$ Then adding $1=\frac{\theta}{\theta}$
$$\frac{1-\theta}{\theta}+\frac{\theta}{\theta}=\overline x\Rightarrow \frac{1-\theta+\theta}{\theta}=\overline x$$
I leave the rest for you. Feel free to ask if something is unclear.