Does the series $\sum \frac{x^n}{1+x^n}$ converge uniformly on $[0,1)$? Consider the following series for $x \in [0,1)$:
$$
\sum \frac{x^n}{1+x^n}
$$
I figured that it converges, by using the ratio test: 
$$
\left| \frac{a_{n+1}}{a_n} \right| = \frac{x^{n+1}}{1+x^{n+1}} \cdot \frac{1+x^{n}}{x^n} = \frac{x+x^{n+1}}{1+x^{n+1}} < \frac{1+x^{n+1}}{1+x^{n+1}} = 1.
$$
with $a_n = \frac{x^n}{1+x^n}$. However, I would like to determine if this convergence is uniform on this interval. I know that for $a<1$ it is uniform on the interval $[0,a]$, since $g_n(x) = \frac{x^n}{1+x^n}$ is increasing on the interval, so we have for every $n$:
$$
|g_n(x)| \leq \frac{a^n}{1+a^n}
$$
and $\sum \frac{a^n}{1+a^n}$ converges, as just shown, so the convergence of $\sum g_k(x)$ is uniform by the Weierstrass M-test. 
However, is it uniform on $[0,1)$?

The convergence is not uniform. For $x_n = \sqrt[n+1]{1- \frac1n}$ we have:
$$
\sup_{x \in [0,1)} \left|\sum_{k=1}^\infty \frac{x^k}{1+x^k} - \sum_{k=1}^n\frac{x^k}{1+x^k}\right| = \sup_{x \in [0,1)} \sum_{k=n+1}^\infty\frac{x^k}{1+x^k}
\ge \frac{x_n^{n+1}}{1+x_n^{n+1}}
= \frac{1-\frac1n}{2-\frac1n} \xrightarrow{n\to\infty} \frac12
$$
so it doesn't converge to $0$.
