Radius of convergence $\sum_{n=1}^{\infty} \frac{x^n}{1+x^n} $ I want to prove that the radius of convergence of the series:
$$\sum_{n=1}^{\infty} \frac{x^n}{1+x^n} $$
is $r=1$. Yeah in this interval does converge by comparison with the geo series $\sum_{n=1}^{\infty} x^n $. 
But how to prove it?
Addendum:
For $x \in (-1, 1)$ we see that the series converges absolutely:
$$\left | \sum_{n=1}^{\infty}\frac{x^n}{1+x^n} \right |\leq \sum_{n=1}^{\infty}\left | \frac{x^n}{1+x^n} \right |\leq \sum_{n=1}^{\infty}|x|^n$$
and for $x>1$ the function inside the series goes to $1$ instead of $0$ hence it does not converge. But how to prove that the same holds for $x<-1$?

If $x<-1$, then $|x|^n \pm 1 > 0$ and $|x|^n \to \infty$ as $n$ grows.
$$f_n(x) \,=\,  \frac{(-|x|)^n}{(-|x|)^n+1} \,=\,  \frac{|x|^n}{|x|^n+(-1)^n} \,>\, 0.$$
$$\frac{1}{f_n(x)} \,=\,   \frac{|x|^n+(-1)^n}{|x|^n}\,=\, 1+  \frac{(-1)^n}{|x|^n} \to 1.$$
Therefore the general term $f_n(x) \to 1$ with $n$, so $\sum f_n(x)$ cannot converge when $x<-1$.
For the absolute convergence when $-1<x<1$, note that $0 \leq |x|^n < 1$ and $|x|^n \to 0$ as $n$ grows. For all large enough $n$,
$$\left |f_n(x)\right | \,=\,  \frac{  |x|^n  }{|1+x^n|} \,\leq\, \frac{  |x|^n  }{1-|x|^n}\,\leq\, \frac{  |x|^n  }{\frac{1}{2}}.$$
We know $\sum 2|x|^n < \infty$. Hence
$$\sum_{n=1}^{\infty} \left | f_n(x)\right | \,=\,\sum_{n=1}^{\infty} \left | \frac{x^n}{1+x^n} \right | < \infty$$
holds also.
