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.
Best Answer
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 $\left|\sin\left(\dfrac x{n^2}\right)\right|\geqslant\dfrac12$.