# Why $\sum_{n=1}^{\infty}\frac{x^{\alpha}}{1+n^2 x^2}$ doesn't converge uniformly on $[0, \infty)$ for $\alpha > 2$?

I'm trying to understand why $$\sum_{n=1}^\infty\frac{x^\alpha}{1+n^2 x^2}$$ doesn't converge uniformly on $$[0, \infty)$$ for $$\alpha > 2$$.

My book says that $$\frac{x^\alpha}{1+n^2 x^2}$$ is monotonic and unbounded for $$\alpha > 2$$ therefore it doesn't converge. I don't get why this means it can't converge exactly, someone care to explain?

• Interesting! Just for fun, the series converges uniformly on $[0, \infty)$ if and only if $1 < \alpha \leq 2$. – Sangchul Lee Jun 28 at 12:23

Because if a series of functions $$\sum_{n=1}^\infty f_n$$ converges uniformly, then the sequence $$(f_n)_{n\in\Bbb N}$$ converge uniformly to the null function. So, the functions $$f_n$$ cannot be unbounded for every $$n\in\Bbb N$$.
• If we fix $x \geq 0$, then for any $\alpha$, we have $\lim_{n \to \infty} \frac{x^{\alpha}}{1+n^2x^2} = 0$. So, this can't be right (unless I'm completely overlooking something) – peek-a-boo Jun 28 at 11:03
When $$\alpha > 2$$, then $$\sup\frac{x^{\alpha}}{1+n^2 x^2} = +\infty$$ so general member doesn't converged to $$0$$ uniformly.