The series $\sum_{k \geq 2} e^{nx}$ is pointwise convergent to its sum $S(x)=\frac{e^{2x}}{1-e^x}$ for all $x \in (-\infty,0)$ because it is a geometric seriesa and $-1<e^{x}<1 \iff x<0$.
I have a question about the uniform convergence: the supremum of the partial sums is $$\sup_{x \in (-\infty,0)} \left|S_n(x)-S(x)\right|=\sup_{x\in(-\infty,0)} \left\{\sum_{k=n+1}^{\infty} e^{kx}\right\}=\sup_{x\in(-\infty,0)} \frac{e^{x(n+1)}}{1-e^x}$$ The function $x \mapsto \frac{e^{x(n+1)}}{1-e^x}$ is increasing for all $n\in\mathbb{N}$ and for all $x\in(-\infty,0)$, and it has a maximum for $x_n=\log\left(\frac{1}{n}+1\right)>0$ for all $n\in\mathbb{N}$; so $$\sup_{x\in(-\infty,0)} \left\{\frac{e^{x(n+1)}}{1-e^x}\right\}=\lim_{x \to 0^-} \frac{e^{x(n+1)}}{1-e^x}=\infty$$ Here comes my doubt: to end the study of the uniform convergence in $(-\infty,0)$, now I have to take the limit as $n\to\infty$ of this supremum, but the supremum is already $\infty$. Intuitively, I would say that the limit of the supremum is again $\infty$ and so the series of functions is not uniformly convergent to $S$ in $(-\infty,0)$, but I'm not sure if this is a correct reasoning and moreover, even if it is correct, I would like to understand rigorously what it means to evaluate a limit of a supremum in the case where the supremum is $\infty$. Is this related to the algebra in $\mathbb{R} \cup \{-\infty,\infty\}$ and so I can treat $\infty$ like a number here or it is something else?
So the question is, what is the correct reasoning in a case like this when I must take the limit of a supremum that is already $\infty$? Thanks.
To conclude the exercise: instead, for every $a>0$ the convergence is uniform in $(-\infty,-a]$ because for the very same reasoning it is
$$\sup_{x\in(-\infty,-a]} \left|S_n(x)-S(x)\right|= \left[\frac{e^{x(n+1)}}{1-e^x}\right]_{x=-a}=\frac{e^{-a(n+1)}}{1-e^{-a}} \to 0 \ \text{when} \ n \to \infty$$
We have uniform convergence if for any $\epsilon > 0$ there exists a positive integer $N$ such that $\left|\sum_{k = n+1}^\infty e^{kx} \right|< \epsilon $ for all $n \geqslant N$ and for all $x \in (-\infty,0)$. This is true if and only if there exists $N$ such that for all $n \geqslant N$,
$$\tag{#}\sup_{x \in (-\infty,0)}\left|\sum_{k = n+1}^\infty e^{kx} \right|\leqslant \epsilon$$
If for any $N$ we can find $n > N$ such that (#) fails to hold, then the convergence is not uniform. You demonstrated an even stronger condition that implies non-uniform convergence. That is, the supremum is infinite for every positive integer $n$. Nothing needs to said about a limit -- although arguing in that way is not incorrect.
Alternatively, if you are uncomfortable with the concept of taking the limit of a sequence where each term is infinite, you could proceed with
$$\sup_{x \in (-\infty,0)}\left|\sum_{k = n+1}^\infty e^{kx} \right| =\sup_{x \in (-\infty,0)}\sum_{k = n+1}^\infty e^{kx}\geqslant \sup_{x \in (-\infty,0)}\sum_{k = n+1}^{2n} e^{kx} \geqslant \sup_{x \in (-\infty,0)}n \cdot e^{2nx} \geqslant n \cdot e^{2n\cdot\frac{1}{n}}= ne^2,$$
Now the limit of the RHS and, consequently, the LHS must be $+\infty$.