When the sum of functions $$\sum_{n=0}^{+\infty} \frac{2^{nx}-1}{4^{nx}+1}$$ convergens pointwisely and uniformly?
I found that this sum converges iff $x\in [0,+\infty)=I$ (otherwise the general term does not go to $0$).
Then, for the uniform convergence I searched the $\sup$ of the general term, so $||a_n (x)||_{\infty, I}$ (the infinity norm in $[0,+\infty)$ interval).
I found that the derivative of $a_n (x)$ is always positive (in $I$) so, $$\sup_{x\in I}|a_n(x)|=\lim_{x\to \infty}|a_n(x)| \sim |\frac{1}{2^n}| \rightarrow 0$$ So there is uniform convergence in I.
But near $x=0$ (when $a_n(x)=0$) do I have uniform convercence?
Uniform convergence of the sequence of general terms is a necessary (but not sufficient) condition for uniform convergence of the series. Nevertheless, it can be a quick way to rule out series uniform convergence.
It is not true that $\sup_{x \in I} |a_n(x)| = \lim_{x \to \infty}|a_n(x)|$. With $n$ fixed, it is true that $|a_n(x)| \to 0$ as $x \to \infty$, but that is irrelevant with regard to uniform convergence.
Note that for all $n \in \mathbb{N}$ we have $1/n \in I = [0,\infty)$ and
$$\sup_{x \in I} |a_n(x)| = \sup_{x \in I} \frac{2^{nx}-1}{4^{nx} + 1} \geqslant \frac{2^{n(1/n)}-1}{4^{n(1/n)} + 1} = \frac{1}{5}. $$
Since the RHS does not converge to $0$ as $n \to \infty$, the general term and, hence, the series fail to converge uniformly on $I$.
Your suspicion was correct that $x \in I$ arbitrarily close to $0$ prevents uniform convergence on $I$.