Is the sum uniformly continuous on $[0,\infty)$?

Is the $$\sum_{n=1}^\infty \frac{nx}{1+n^3x^2}$$ uniformly convergent on $$(0,\infty)$$?

Each individual term attains a max of $$\frac{1}{2n^{1/2}}$$ at $$x=\frac{1}{n^{1.5}}$$.

Notice that the maximizers move to the left and maximums shrink as $$n$$ increases. So, one cannot put a lower bound of $$\sum \frac{1}{2n^{0.5}}$$ in a crude way. But I believe with some more algebra we might still bound from below by the latter sum which diverges. And that is what I am asking!

• Consider $g(x):=x^{-2/3}\in L^1[0,1]$. For your series $\sum_{k\ge1}f_k(x)$, it is easy to check $\int_0^1 f_k g\, dx\ge {1\over2k}$. Therefore $\int_0^1\big(\sum_{k\ge1}f_k\big)g\, dx=+\infty$. So $\sum_{k\ge1}f_k$ is not bounded on $[0,1]$ and the convergence is not uniform. – Pietro Majer Nov 9 '19 at 14:08