Let $X_1, X_2, . . . , X_n$ be iid random variables having pdf
$$f_X(x\mid\theta) =\theta^{2}(1 -\theta)^{x}$$
where $$x\in\{ 0,1,... \}$$
I want test this hypothesis using UMP test:
$$H_0:\theta=\theta_0\space vs \space H_1:\theta >\theta_1$$
By Neyman–Pearson lemma where the test is of the type and the pdf is $f(x|\theta)$ with critical region $R$ is $x\in R$ if $\frac{f(x|\theta_1)}{f(x|\theta_0)}>k$ and $x\in R^c$ if $\frac{f(x|\theta_1)}{f(x|\theta_0)}<k$ for any $k\geq 0$ and $\alpha=P_{\theta_0}(X\in R)$
$$ \frac{f(x|\theta_1)}{f(x|\theta_0)}>k$$ $$\frac{\theta_{1}^{2n} (1-\theta_{1})^{\sum_{i=1}^{n} x_{i}} }{\theta_{0}^{2n} (1-\theta_{0})^{\sum_{i=1}^{n} x_{i}}}>k$$
$$\left(\frac{\theta_{1}}{\theta_{0}} \right)^{2n} \, \left(\frac{1-\theta_{1}}{1-\theta_{0}} \right)^{\sum_{i=1}^{n} x_{i}} > k $$
Therefore, $$ {\sum_{i=1}^{n} x_{i} < k_{1}}$$
Let us assume that I wanted to find the power of $\theta$ using the normal approximation. At this point, I do not know what is the distribution of $X$. If I were to find the expectation and variance of the above distribution it will be somehow complicated i.e $$ \mathbb{E}(X)= \sum_{x=0}^{\infty} \theta^{2} x (1-\theta)^{x}=1-\theta \, \, \text{when} \, \, |1 - \theta|<1$$
$$\mathbb{E}(X^2) = \sum_{x=0}^{\infty} \theta^2 x^2 (1 - \theta)^x = \frac{(\theta^2 - 3 \theta + 2)}{\theta} \, \, \text{when} \, \, |1 - \theta|<1$$
But I was hoping if it belongs to a well-known distribution or that there is a better approach to this problem.