I tried to solve this problem. Much of the obvious things to try are there but don't work. After several failed attempts I arrived at this (also failed?) argument:
Let $$f_n(x)=\frac{nx}{1+n^4x^2}$$ The goal is to prove that $$F_m(x)=\sum_{n=0}^{m}(-1)^nf_n(x)$$ converges uniformly in $[0,\infty)$. I've tried the following:
First of all, for every $n\in\mathbb{N}$ and every $x\in[0,\infty)$ we have $|f_n(x)|<\frac{1}{2n}$
Given $x_0\in[0,\infty)$ and taking $f_n(x_0)$ as a function of $n$ we get that the sequence $\{f_n(x)\}_{n\in\mathbb{N}}$ starts to be decrasing when $n^4\ge\frac{1}{3x_0^2}$ or $x_0\ge\frac{1}{n^2\sqrt{3}}$. Now, for $p>q$ $$|F_p(x_0)-F_q(x_0)|=\left|\sum_{n=q+1}^{p}(-1)^nf_n(x_0)\right|$$ and we separate the following cases:
If $x_0\ge\frac{1}{(q+1)^2\sqrt{3}}$ then $\{f_n(x_0)\}_{n\ge q+1}$ is decreasing so we can use "Leinbiz's criterion proof" to get $$|F_p(x_0)-F_q(x_0)|<\left|f_{q+1}(x_0)\right|<\frac{1}{2(q+1)}$$ If $x_0\le\frac{1}{p^3\sqrt{3}}$ then $x_0<\frac{1}{n^3}$ for every $n\le p$ and we have $$|F_p(x_0)-F_q(x_0)|=\left|\sum_{n=q+1}^{p}(-1)^nf_n(x_0)\right|<\sum_{n=q+1}^p\frac{1}{n^2}<\frac{1}{q}$$ which we can make small.
If $\frac{1}{p^3\sqrt{3}}\le x_0\le\frac{1}{(q+1)^2\sqrt{3}}$ then none of the bounds i came up with tend to zero. Does someone have an idea? If this worked somehow I think it would be a neat thing to try for other problems.
Edit: Please note that when I uploaded my question, the original one already had an accepted answer. I'm not looking for a solution of that exercise but rather for some way to complete my line of reasoning. The functions $f_n$ or $F_m$ don't interest me as much as the "dividing the real line" trick and if it works here or in general for alternating sums of functions
Thanks!