Does the series $\sum \frac{x^n}{1+x^n}$ converge uniformly on $x\in [0,1)$? Does the series $\sum \frac{x^n}{1+x^n}$ converge uniformly on $x\in [0,1)$?
I have no idea where to start. Could somebody give me  a hint ?
Edit: Could I use something like this ?
$$\left|\frac{x^n}{1+x^n}\right|\leq x^n$$
Because $x<1$, the geometric series $\sum x^n$ converges. Therefore by the M-test we get that the series converges uniformly.
 A: The partial sums $S_N(x)=\sum_{n=0}^N \frac{x^n}{1+x^n}$ converge pointwise on $[0,1)$ to $S(x)=\sum_{n\geq 0} \frac{x^n}{1+x^n}$ by comparison with the geometric series $\sum x^n$.
Now
$$
 S(x)-S_N(x)=\sum_{n\geq N+1} \frac{x^n}{1+x^n}\geq \frac{x^{N+1}}{1+x^{N+1}}
$$
for all $x\in[0,1)$.
So, letting $x$ tend to $1$, we get
$$
\sup_{[0,1)}S-S_N\geq \frac{1}{2}
$$
for all $N$.
Hence the convergence to $S$ is not uniform (which by definition is $\sup_{[0,1)}|S-S_N|\longrightarrow 0$ as $N\rightarrow +\infty$.)
Note: you don't even need $S$, you could simply do it with the Cauchy criterion, if you use the fact that $C([0,1),\mathbb{R})$ is complete when equipped with the uniform norm.
Then
$$
|S_M(x)-S_N(x)|\geq \frac{x^M}{1+x^M}
$$
for all $M>N$ and all $x\in[0,1)$. 
Hence
$$
\sup_{[0,1)}|S_M-S_N|\geq \frac{1}{2}
$$
and the sequence is not uniformly Cauchy.
A: By definition the serie $\displaystyle\sum\frac{x^n}{1+x^n}$ converges uniformly on $[0,1)$ if 
$$\lim_n\sup_{x\in[0,1)} \sum_{k=n+1}^{\infty}\frac{x^k}{1+x^k}=0.$$
We have 
$$\sum_{k=n+1}^{\infty}\frac{x^k}{1+x^k}\geq \sum_{k=n+1}^{\infty}\frac{x^k}{2}=\frac{1}{2}\frac{x^{n+1}}{1-x},$$
So it's clear that $\displaystyle\sup_{x\in[0,1)} \frac{x^{n+1}}{1-x}=+\infty, $ and hence the condition of uniform convergence is not verified.
