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I asked this question about a week ago but I am little bit unsure about the way to solve it so I hope it is ok if I ask again about some things I do not fully understand.

I have to show that the series $$ S = \sum_{n=1}^\infty \sin \left( \frac{x}{n^2} \right) $$ does not converge uniformly on $\mathbb{R}$ which can be shown by showing that $ \sin \left( \frac{x}{n^2} \right) $ fails to uniformly converge towards $0$ when $n$ tends to $\infty$. Is this because of contraposition? I know that if $\sum_{n=1}^\infty a_n$ converges unifomrly then $a_n$ converges uniformly towards $0$.

Futhermore, by negation we have that $\sin \left( \frac{x}{n^2} \right)$ does not converge uniformly towards $0$ when $n$ tends to $\infty$ if $$ \exists \epsilon > 0 \ \forall N \in \mathbb{N} \ \exists x \in \mathbb{R} \ \exists n \in \mathbb{N} \ : n \geq N \ \text{and} \left|\sin \left( \frac{x}{n^2} \right)\right| \geq \epsilon $$ If I then pick $\epsilon = \frac{1}{2}$ and $x = \frac{\pi n^2}{3}$ I get the desired result but don't I also have to pick a specific $n \in \mathbb{N}$ so that this only works when $n \geq N$? Or is it simply enough to pick $\epsilon$ and $x$?

Thanks for your help.

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    $\begingroup$ The series of functions is Uniformly convergent at any $[a,b]$. $\endgroup$ Jun 1, 2020 at 21:50
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    $\begingroup$ I assume this was your earlier question $\endgroup$
    – Henry
    Jun 1, 2020 at 21:54
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    $\begingroup$ As stated, the question makes no sense. It doesn't converge uniformly where? $\endgroup$ Jun 1, 2020 at 21:56
  • $\begingroup$ I am sorry. I have edited now. I have to show it does not converge uniformly on $\mathbb{R}$. $\endgroup$
    – Mathias
    Jun 1, 2020 at 21:58
  • $\begingroup$ I would expect you to pick $x$ and $n$ as functions of $\epsilon$ and $N$ $\endgroup$
    – Henry
    Jun 1, 2020 at 21:59

1 Answer 1

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Concerning your first question: yes, it is by contraposition.

The proof that your sequence does not converge uniformly to $0$ is almost correct. Simply take $x=\dfrac{\pi N^2}3$ (instead of $x=\dfrac{\pi n^2}3$). Then, there is a $\varepsilon>0$ (namely, $\dfrac12$) such that, for every $N\in\Bbb N$, there is some natural $n\geqslant N$ (namely, $N$ itself) and some number $x$ such that $\left|\sin\left(\dfrac x{n^2}\right)\right|\geqslant\dfrac12$.

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  • $\begingroup$ Ok I think I understand now. Thanks for your time once again, Jose. $\endgroup$
    – Mathias
    Jun 1, 2020 at 22:09
  • $\begingroup$ Are you always able to pick the natural number $n$ to be $N$? such that $n \geq N$ or does it only work in some scenarios? $\endgroup$
    – Mathias
    Jun 1, 2020 at 22:13
  • $\begingroup$ @Mathias Why? If $x=\frac{\pi N^2}3$, then $\left|\sin\left(\frac x{N^2}\right)\right|=\frac12$. $\endgroup$ Jun 1, 2020 at 22:16
  • $\begingroup$ Now I think I got confused again. I thought if we picked $\frac{\pi n^2}{3}$ (which was not correct I know) one would have that $\left|\sin( \frac{\pi n^2}{3n^2})\right| = \frac{\sqrt{3}}{2} \geq \frac{1}{2}$ but I don't see how this is the same as what you just wrote? $\endgroup$
    – Mathias
    Jun 1, 2020 at 22:20
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    $\begingroup$ Yes, that is correct. $\endgroup$ Jun 1, 2020 at 22:33

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