# UMP to Geomtric distribution

Given the hyphotesis

$$H_0 : \theta_0 \space vs\space H_1 : \theta_1$$

I want find the Uniformely Most Powerful test (UMP).

Well, we know by the Neyman-Pearson lemma that: $$\frac{f(\widetilde{x}/\theta_1)}{f(\widetilde{x}/\theta_0)} > k$$ to some $$k > 0$$ we will have a UMP with size $$\alpha$$ such that $$P(X \space \epsilon \space R | \theta = \theta_0) = \alpha$$

Doing this to the Geometric distribution I will find something like this:

$$\sum{X_i} \geq \frac{k + n ln(\frac{\theta_1}{\theta_0})}{ln(\frac{1 - \theta_1}{1 - \theta_0})}$$

My doubt is how to proced by here. I know that $$\frac{k + n ln(\frac{\theta_1}{\theta_0})}{ln(\frac{1 - \theta_1}{1 - \theta_0})} = c$$ such that

$$P(\sum{X_i} \geq c) = \alpha$$ but how can I sum all the geometric distribution? Is this new distribution a normal or another one?