Decide whether $\sum_{n=1}^{\infty}(-1)^n\frac{nx}{1+n^4x^2}$ uniformly converges on $[0,\infty)$ or not

$\newcommand{\FUNC}[1]{#1 \frac{nx}{1+n^4x^2}}$

The following problem has bothered me for a very long time. I've been trying to solve this problem for 3 days long - and it feels I tried everything. To conclude - I absolutely need the help of the professionals; And that's why I came here.

I am given with the following function series:

$$S(x)\equiv\sum_{n=1}^{\infty}(-1)^nf_n(x)\equiv\sum_{n=1}^{\infty}\FUNC{(-1)^n}$$

My task is to decide whether the series $S(x)$ uniformly converges on $[0,\infty)$, or not.

Things I tried:

(1) I tried to use Leibniz's Alternating Series Test. I proved that $f_n(x)$ uniformly converges to $0$, but the problem is that $f_n(x)$ is not a decreasing sequence - for every $n\in\mathbb{N}$, there exists $0<x_0\in\mathbb{R}$, such that $f_n(x)$ is increasing on $[0,x_0]$, and decreasing on $[x_0,\infty)$. I've noticed that $\lim_{n\to\infty}x_0=0$, but that doesn't help - since it would never actually reach $0$ (that would only help to prove that the series uniformly converges on $[c,\infty)$ when $c>0$).

(2) I tried to use Weierstrass's M Test. I found that $f_n(x)\leq\frac{1}{2n}$ for every $n\in\mathbb{N}$, but that of course won't help (since $\sum_{n=1}^{\infty}\frac{1}{2n}$ diverges). The huge problem using the M test is that $\frac{1}{2n}$ is actually a tight bound.

(3) I proved that $\displaystyle \sum_{n=1}^{\infty}\FUNC{}$ uniformly converges on $[c,\infty)$ when $c>0$, but does not uniformly converges on $[0,\infty)$. That wasn't very helpful, since $S(x)$ might converges conditionally.

(4) Defining:

$$S_m(x)\equiv\sum_{n=1}^{m}\FUNC{(-1)^n}$$

It is sufficient to show that:

$$(*) \lim_{m\to\infty}\left(\sup_{[0,\infty)}\left|S_m(x)-S(x)\right|\right)\equiv\lim_{m\to\infty}\left(\sup_{[0,\infty)}\left|\sum_{n=m+1}^{\infty}\FUNC{(-1)^n}\right|\right)=0$$

in order to prove the series uniformly converges on $[0,\infty).$ Looks impossible and complicated - yet I was eager to try that too. But to no avail.

I will say that my intuition is that the series does uniformly converge on $[0,\infty)$, and that is because I used graphic calculators, and found that for every $n\in\mathbb{N}$:

$$\sup_{[0,\infty)}\left|\sum_{n=m+1}^{\infty}\FUNC{(-1)^n}\right|\leq\sum_{n=m+1}^{\infty}\frac{1}{n^2}$$

And since $\displaystyle \sum_{n=m+1}^{\infty}\frac{1}{n^2}$ converges, we can say that its limit when $m\to\infty$ is $0$, thus proving $(*)$. However - I couldn't prove this inequality. I tried using the triangle inequality, to distinguish between odd and even indices - nothing worked.

As you might see - I really need help. Thank you very much!

The following estimate will be useful:

Claim. For any $a, b, x \geq 0$, we have

$$ \left| \frac{bx}{1+b^4x^2} - \frac{ax}{1+a^4x^2} \right| \leq \frac{3x|b-a|}{1+\min\{a,b\}^4x^2}. $$

Proof of Claim. Assume $a < b$ without losing the generality. Then by the fundamental theorem of calculus,

\begin{align*} \left|\frac{bx}{1+b^4x^2} - \frac{ax}{1+a^4x^2}\right| &\leq \int_{a}^{b} \left|\frac{\partial}{\partial \alpha} \left( \frac{\alpha x}{1+\alpha^4x^2} \right) \right| \, \mathrm{d}\alpha = \int_{a}^{b} \left| \frac{x(1-3\alpha^4 x^2)}{(1+\alpha^4x^2)^2} \right| \, \mathrm{d}\alpha. \end{align*}

Now using the triangle inequality, $\left|1-3\alpha^4 x^2\right| \leq 1+3\alpha^4x^2 \leq 3(1+\alpha^4x^2)$, and so, the last integral can be bounded from above by

\begin{align*} &\leq \int_{a}^{b} \frac{3x}{1+\alpha^4x^2} \, \mathrm{d}\alpha \leq (b-a) \max_{a \leq \alpha \leq b} \left(\frac{3x}{1+\alpha^4x^2}\right). \end{align*}

Since the function $\alpha \mapsto 3x/(1+\alpha^4x^2)$ is decreasing in $\alpha$, the maximum is achieved at $\alpha = a$ and the desired inequality follows. ////

Returning to the original problem, note that

$$ \left| f_n(x) - f_{n+1}(x) \right| \leq \frac{3x}{1+n^4x^2} \leq \frac{3}{2n^2}, $$

where the first step is a consequence of the claim and the second step is simply an application of the AM-GM inequality. Therefore, by the Weierstrass M-test, the sum

$$ \sum_{k=1}^{n} (f_{2k-1}(x) - f_{2k}(x)) $$

converges uniformly on $[0, \infty)$, from which the uniform convergence of $S(x)$ easily follows.