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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?

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

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    $\begingroup$ 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) $\endgroup$ – peek-a-boo Jun 28 at 11:03
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    $\begingroup$ @peek-a-boo My bad. I've edited my answer. $\endgroup$ – José Carlos Santos Jun 28 at 11:09
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When $\alpha > 2$, then $\sup\frac{x^{\alpha}}{1+n^2 x^2} = +\infty$ so general member doesn't converged to $0$ uniformly.

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