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 $[0,1)$?

No, it is not uniform.
Set $S_n(x)=\sum_{k=1}^{n}\frac{x^k}{1+x^k}$.
Then, for $n>m$,
$$\sup_{x\in[0,1)}|S_n(x)-S_m(x)|\geq\sum_{k=m+1}^{n}\frac{1^k}{1+1^k}=\frac{n-m}{2}$$
Therefore $$\lim_{n\to\infty}\sup_{x\in[0,1)}|S_n(x)-S_m(x)|$$ is not finite, and so $S_n$ is not uniformly Cauchy, i.e the series does not converge uniformly.
