Discuss the uniform convergence of this series over $\mathbb R$: $$\sum_{n=1}^\infty\frac{(-1)^n x^n}{n(1+x^n)}.$$
For $0<x\leq 1$, $\frac{x^n}{n(1+x^n)}$ is monotonically decreasing and coverges to zero as $n$ tends to $\infty$.
Similarly for $x > 1$, $\frac{x^n}{n(1+x^n)}$ monotonically decreses for $x$ and converges to zero.
In both case they are bounded uniformly between $0$ and $1$, and $\sum_n{(-1)^n}$ oscillates finately between $0,1$ as $n$ is even or odd, by Abel Test we conclude that series is Uniformly convergence over $(0,+\infty)$.
Doubt what about when $x \leq 0$, I am confused as I am getting alternating series i.e. $$\frac{-x}{1+x}$ + $\frac{x^2}{2(1+x^2)}$-$\frac{x^3}{3(1+x^³)}\dots$$ Here as $x$ tends to $-1$, denominator turns $0$. Please help.
The series is not defined at $x = -1$. Furthermore we can show that it even fails to converge uniformly on $(-1,0]$ when this point is excluded.
Let $y = -x$ for $x \in (-1,0]$ and consider the equivalent series $\displaystyle \sum_{n=1}^\infty\frac{y^n}{n(1+(-1)^ny^n)}$ for $y \in [0,1)$.
We have $y_n = 1 - 1/n \in [0,1)$ for all $n$ and
$$\sup_{y \in [0,1)}\left|\sum_{k=n+1}^\infty\frac{y^k}{k(1+(-1)^ky^k)} \right| \geqslant \sup_{y \in [0,1)}\sum_{k=n+1}^{2n}\frac{y^k}{k(1+(-1)^ky^k)} \geqslant \sup_{y \in [0,1)}n \cdot \frac{y^{2n}}{2n(1+y^n)} \\ \geqslant n \cdot \frac{(1-1/n)^{2n}}{2n(1+(1-1/n)^n)} = \frac{(1-1/n)^{2n}}{2(1+(1-1/n)^n)} \underset{n \to \infty}\longrightarrow \frac{e^{-2}}{2(1+e^{-1})} \neq 0,$$
Thus, the series fails to converge uniformly by violation of the necessary condition
$$\lim_{n \to \infty}\sup_{y \in [0,1)}\left|\sum_{k=n+1}^\infty\frac{y^n}{k(1+(-1)^ky^k)} \right|= 0$$