Does the series $\sum \frac{x^n}{1+x^n}$ converge uniformly on $[0,1)$?

Question

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 $[0,1)$?

Answer 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.

Answer 2

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 $0$.

Answer 3

In general, if $\sum f_n$ converges uniformly on a set $E,$ then $\sup_E |f_n| \to 0.$ In our problem we have

$$\sup_{[0,1)} \frac{x^n}{1+x^n} \ge \sup_{[0,1)} \frac{x^n}{2} = \frac{1}{2}.$$

Thus the series fails to converge uniformly.