My problem: To determine and prove whether the series $\sum \frac{x^{n}}{1+x^{n}}$ converges uniformly on $ [0,1)$.
My solution: First, I should note that despite seeing graphically that it looks like the series might converge uniformly on $[0,1)$, my intuition tends towards the opposite.
Claim. The series $\sum \frac{x^{n}}{1+x^{n}}$ does not converge uniformly on $ [0,1)$.
Proof. First note, that convergence of the series means convergence of the sequence of its partial sums; this means that $\sum \frac{x^{n}}{1+x^{n}}$ converges uniformly on $[0,1)$ if and only if so does the sequence $\displaystyle (g_n (x)) = \left(\sum_{k=1}^n \frac{x^{k}}{1+x^{k}} \right) $. Additionally, $f_n$ is continuous for all $n \in \mathbb{N}$ and for all $x\in [0,1)$, which implies that the limit function must also be continuous (by the Theorem 24.3). In other words, the function \begin{equation}\nonumber \lim_{n \to \infty} g_n(x) = \sum\frac{x^n}{1+x^n} = g(x) \end{equation} must be continuous on $[0,1)$. However, we know that for $x=0$ the series converges to $0$, while for $x\neq 0$ $\sum \frac{x^{n}}{1+x^{n}}>0$. (There is a theorem in our book regarding the uniform continuous limit of a function being a continuous function which is what I was trying to go into.)
In the last bit, I mean to somehow say that the series loses its $n$-dependency and therefore is not continuous. I'm not sure how to. Also, I am not sure whether it is a legitimate reason for lacking continuity.
Any thoughts?