I'm analysing the series of functions $$\sum_{n=1}^{\infty} \frac{1}{e^{x\cdot n^2}-1} \qquad \textrm{where } \space x \geq 0$$ for uniform convergence. I'm trying to use the Weierstrass M-test so I picked a convergent upper bound of $\displaystyle{ \frac{1}{n^2} }$ which holds when $x\geq 1$, i.e: $$\left|\frac{1}{e^{n^2x}-1}\right| \leq \left|\frac{1}{n^2}\right| \space \textrm{for } x\geq1, $$ and we know that $$\zeta (2) = \sum_{n \mathop = 1}^{\infty} {\frac 1 {n^2}} = \frac {\pi^2} {6} < \infty$$
However, when $x \in [0;1)$, I'm unsure since the value is undefined when $x=0$. Does this imply that the series of functions is thus not uniformly convergent for $x \geq 0$, but is uniformly convergent for $x \gt 0$? Or are we supposed to only focus only on what happens in the limit?
Hint: If $\sum f_n$ converges uniformly on $X,$ then $\sup_X |f_n|\to 0.$ Is that true for the given series on $X=(0,\infty)?$?