Uniform convergence of $\sum_{n=1}^{\infty}\frac{x^n}{1+x^n}$

Question

Where does the seriers $\displaystyle \sum_{n=1}^{\infty}\frac{x^n}{1+x^n}$ converges ? On which interval does it converge uniformly ?

We have $$\lim_n \frac{x^n}{1+x^n}=\begin{cases}0 &\text{ , if } |x|<1\\\frac{1}{2} &\text{ , if } x=1\\\text{does not exists} &\text{ , if } x=-1\\1 &\text{ , if } |x|>1\end{cases}$$ So we have to test the convergence only on the interval $(-1,1)$.

Now let , $\displaystyle S_n=\sum_{k=1}^n \frac{x^k}{1+x^k}$. How I find the point wise sum ?

Edit: I'm confused for different answers and different comments..!!

Answer 1

Observe for all $|x| \leq R<1$ you have that \begin{align} \left|\sum^m_{n=k} \frac{x^n}{1+x^n}\right| \leq \sum^m_{n=k} \frac{|x|^n}{1-R^n} \leq \frac{1}{1-R}\sum^m_{n=k}R^n = \frac{R^k-R^{m+k+1}}{(1-R)^2} \end{align} which means the series is uniformly Cauchy on $[-R, R]$. Hence the series is uniformly convergent on any compact subset of $(-1, 1)$.

Edit: Suppose the series is uniformly convergent on all of $(-1, 1)$. Fix $\epsilon = \frac{1}{2}$, then there exists $N$ such that \begin{align} \left|\sum^\infty_{n=N}\frac{x^n}{1+x^n}\right| <\frac{1}{2} \end{align} for all $x \in (-1, 1)$. Let $x_\ell = \frac{\ell-1}{\ell}$, then observe that \begin{align} \sum^\infty_{n=N} \frac{x_\ell^n}{1+x_\ell^n} = \sum^\infty_{n=N} \frac{(\ell-1)^n}{\ell^n+(\ell-1)^n} \geq \frac{1}{2}\sum^\infty_{n=N}\left(\frac{\ell-1}{\ell}\right)^n = \frac{1}{2}\frac{\left(\frac{\ell-1}{\ell}\right)^N}{1-\frac{\ell-1}{\ell}} = \frac{\ell}{2}\left(1-\frac{1}{\ell} \right)^N \rightarrow \infty \end{align} as $\ell\rightarrow \infty$.

Answer 2

A NT-point of view related with Lambert series: assuming $|x|<1$ we have $$ \frac{x^n}{1+x^n} = x^{n}-x^{2n}+x^{3n}-\ldots \tag{1}$$ hence $$ \sum_{n\geq 1}\frac{x^n}{1+x^n} = \sum_{n\geq 1}\delta(n) x^n\tag{2} $$ where $\delta(n)$ is the difference between the number of odd divisors of $n$ and the number of even divisors of $n$. If we represent every positive integer $n$ as $2^{\nu_2(n)}\cdot m$ (so that $m$ is odd) we have $$ \delta(n) = \tau(m)(1-\nu_2(n)) \tag{3} $$ and through a crude bound $$ \left|\delta(n)\right|\leq \sqrt{n}\log_2(n)\tag{4} $$ we get an inequality ensuring the absolute convergence of the RHS (or LHS) of $(2)$ in any compact subset of $(-1,1)$.