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

Solution 1:

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.