Last week I had an assignment to show $f(x)=\sum\limits_{n=1}^\infty\frac{nx^2}{n^3+x^3}$ for $x\ge0$ does not converge uniformly, but I misread the question as "show $f(x)$ is not uniformly continuous."
The actual problem went on to show that $f(x)$ is continuous, but I have been stumped by the question I misread:
Is $f(x)=\sum\limits_{n=1}^\infty\frac{nx^2}{n^3+x^3}$ uniformly continuous on $[0,\infty)$?
I asked my professor about the problem today, but unfortunately we still didn't come up with an answer (and while my professor believes that $f(x)$ is not uniformly continuous, I suspect that it is).
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Things I have proven about $f(x)$ that I can explain or one can assume in an answer: The series does not converge uniformly to $f(x)$, $f(x)$ is continuous, and $f(x)>\frac{x-1}{2}$.
I think that the derivative of $f$ is indeed bounded on $[0,\infty),$ which implies $f$ is uniformly continuous there. I'll give an outline: Let's write
$$f(x) = x^2 \sum_{n=1}^{\infty} \frac{n}{n^3 + x^3}.$$
This will give
$$\tag 1 f'(x) = 2x \sum_{n=1}^{\infty} \frac{n}{n^3 + x^3} + x^2 \sum_{n=1}^{\infty} \frac{-3x^2n}{(n^3 + x^3)^2}.$$
(You verify this on any bounded interval, where all convergence in sight is uniform. Since differentiation is a local property, we get $(1)$.)
Now the right side of $(1)$ will be bounded if we show
$$\sum_{n=1}^{\infty}\frac{n}{n^3 + x^3} = O(1/x) \,\,\text { and } \sum_{n=1}^{\infty}\frac{n}{(n^3 + x^3)^2} = O(1/x^4)$$
as $x\to \infty.$ OK, I'll leave it here for now. Some things to check.