<s>[INST] How to prove that $\sum_1^{\infty} \frac{nx^2}{n^3 + x^3}$ converges my goal here is: 

To prove that \begin{align}f(x) =  \sum_{n=1}^{\infty} \frac{nx^2}{n^3 + x^3} \end{align} converge pointwise on $[0, \infty)$.

Because what I really want is to prove that $\forall A> 0$,the serie on the interval $[0,A]$ converge uniformly (if serie converge pointwise then converge uniformly on a compact interval).


*

*So this is the best way to do that?

*How can I prove that series converges pointwise on $[0, \infty)$?
I have trying to do this too:

\begin{align} f'(x) = \sum \frac{n(2n^3 - x^4)}{(n^3 + x^3)^2} \leq \sum\frac{n^4}{n^6} = \sum \frac{1}{n^2} \end{align}

And this last series is convergent, so I can conclude that the derivative of f is bounded, then f is Lipschitz. But this let me conclude what?
 [/INST] Another (overly complicated in this case) proof would be using that $\int_1^\infty {\rm d}n \frac{n x^2}{n^3+x^3}$ is finite for $x\ge 0$ which implies that $\sum_{n=1}^n \frac{n x^2}{n^3+x^3}$ is finite for $x\ge 0$ as well. To show that the integral is finite, calculate partial fractions first:
$$ \frac{n x^2}{n^3+x^3} = \frac{2 n x - n^2}{3 (n^2 - n x + x^2)} + \frac{n}{3 (n + x)}. $$
Integrating this over n from $1$ to $\infty$ yields the finite result for $x\ge 0$:
$$\frac{1}{6}x \left( 2\,\ln  \left( x+1 \right) -\ln  \left( -x+{x}^{2}+1
 \right) +2\,\sqrt {3}\arctan \left( {\frac {x-2}{\sqrt{3} x}} \right) +\pi \,\sqrt {3} \right).$$
</s>