# Show that the series $\sum_{n=1}^\infty \sin \left( \frac{x}{n^2} \right)$ does not converge uniformly

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 $$x$$?

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

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 $$\left|\sin\left(\dfrac x{n^2}\right)\right|\geqslant\dfrac12$$.
• 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? Jun 1, 2020 at 22:13
• @Mathias Why? If $x=\frac{\pi N^2}3$, then $\left|\sin\left(\frac x{N^2}\right)\right|=\frac12$. Jun 1, 2020 at 22:16
• 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? Jun 1, 2020 at 22:20