How the interval [a, b]: $x \in [a,b]$ can be found for the next sum?
$$\sum_{n=1}^{\infty} \frac{1+x^n}{1-x^n}$$
The sence to check the next limit $$\lim_{n \to \infty} \frac{1+x^n}{1-x^n} = f(x) \ne 0$$ $$f(x) = \begin{cases} 1 , |x|<1 \\ -1 , |x|>1 \\ \emptyset , x = \pm 1 \end{cases}$$ is to define a kind of convergence. $$\lim_{n \to \infty} {\sup_X { | \frac{1+x^n}{1-x^n}-f(x)|}} = s \ne 0$$ $$s = \begin{cases} 2 , |x|<1 \\ -2 , |x|>1 \\ \emptyset , x =\pm 1 \end{cases}$$ It is the non-uniformly convergent series. Am I right?
How can I find the the convergent interval of x? Or prove, that interval = $\emptyset$?
You used L'Hospital rules to calculate the limit. You can only do this if you have a limit of type $\frac00$ or $\frac\infty\infty$. You don't have that if $|x|<1$.
So, for $|x|>1$, your calculation of the limit is correct. Because the limit is not $0$, this tells you a lot about the convergence of the sum.
For $|x|=1$, the sum is not even well defined. Now, analyze what happens if $|x|<1$. hint: calculating the same limit as before may still help.