# Uniform convergence of $\sum e^{-xn^2}$

Good day, I have a question to complete which asks me to prove whether the following series is point-wise convergent and uniformly convergent. Obviously if it does not converge point-wise it will not converge uniformly (using the "strength" of uniform convergence).

$$\sum_{n=0}^\infty e^{-xn^2} \text{ on domain } (0,\infty)$$

I have shown that is is point-wise convergent by comparing it to the Geometric series $$\frac{1}{1-e^{-x}}$$

But I am not quite sure how I can go about disproving uniform convergence. I am tempted to use a corollary of $$\ f_n \to f \text{ uniformly } iff\ \forall\epsilon>0 \ \exists N \ \forall n,m \geq N \ \ \Vert f_n-f_m \Vert_{sup}<\epsilon$$ which states that if $$\ \Vert g_n \Vert_{sup} \nrightarrow 0 \ \implies \sum_{n=0}^\infty g_n \text{ does not converge uniformly}$$

But I am not too sure if that is valid.

It's not uniformly Cauchy. For $M<N$ we have $$\sup_{0<x}\sum_{n=M}^N e^{-n^2x}>\sup_{0<x}e^{-M^2x}=1$$
Whenever $\:\:x\equiv x_n=n^{-2}\in\left(0,\infty\right)=\mathbb R_+,\:$ we have that $$F(x_n)=\lim_{n\to\infty}F_n(x_n)=\lim_{n\to\infty}\sum_{j=0}^ne^{-1}=\infty$$