I'm unable to prove if the following series converge uniformly on $[0,+\infty)$ for the values of the parameter $\beta$ between $2$ and $3$.
$$ \sum_{n\geq 1}\frac{n^\beta x}{x^4+n^4} $$ The maximum of the summed functions is realized (for fixed $n$) in $\frac{n}{\sqrt[4]{3}}$ and it is $\frac{\sqrt[4]{27}}{4}n^{\beta-3}$ Clearly there is total convergence for $\beta<2$ and there is no convergence if $\beta>3$ except in $x=0$. For $2<\beta<3$ pointwise convergence holds but not total. How to prove uniform convergence?
Any hint will be appreciated.
Note that for $x \in [0,\infty)$,
$$\sup_{x \in [0,\infty)}\left|\sum_{k = n+1}^\infty\frac{k^\beta x}{x^4+k^4}\right|\geqslant \sup_{x \in [0,\infty)}\sum_{k = n+1}^{2n}\frac{k^\beta x}{x^4+k^4}\geqslant\sup_{x \in [0,\infty)} n \cdot \frac{n^\beta x}{x^4 + (2n)^4}\\ \underbrace{\geqslant}_{x = n\in [0,\infty)} n \cdot \frac{n^\beta n}{n^4 + (2n)^4}= \frac{n^{\beta+2}}{17 n^4}= \frac{n^{\beta-2}}{17}$$
Since the RHS does not converge to $0$ as $n \to \infty$ for $\beta \geqslant 2$, the series is not uniformly convergent on $[0,\infty)$ for $2 < \beta < 3$.