Uniform convergence of $\sum_{n = 1}^\infty x \sin \frac{1}{(x n)^2}$ on $(0, 1)$. Functional sequence $x \sin \frac{1}{(x n)^2} {}^\longrightarrow_\longrightarrow 0$ because
$$
\left| x \sin \frac{1}{(x n)^2} \right| \le |x| \sqrt{\left|\sin \frac{1}{(x n)^2}\right|} \le |x| \sqrt{\frac{1}{(x n)^2}} = \frac1n \to 0.
$$
It is easy to see that the series $\sum_{n = 1}^\infty x \sin \frac{1}{(x n)^2}$ converges pointwise on $(0, 1)$:
$$
\left| \sum_{n = 1}^\infty x \sin \frac{1}{(x n)^2} \right| \le \sum_{n = 1}^\infty \left| x \sin \frac{1}{(x n)^2} \right| \le |x| \sum_{n = 1}^\infty \frac{1}{n^2 x^2} = \frac{1}{x} \sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6 x}.
$$
But whether the series converges uniformly on $(0, 1)$?
 A: A necessary and sufficient condition for uniform convergence of a series $\sum f_n(x)$ for $x \in  (a,b)$ is the Cauchy criterion.  That is, for all $\epsilon > 0$ there exists $N \in \mathbb{N}$, independent of $x$, such that for all $m > n > N$ and all $x \in (a,b)$ we have $\left|\sum_{k=n+1}^m f_k(x) \right| < \epsilon$.
In this case, we can show that the series is not uniformly convergent on $(0,1)$ by violation of the Cauchy criterion. For any $N$, take $n = \lceil \sqrt{2}N\rceil$, $m = 2N$, and $x_N = \frac{1}{N \sqrt{\pi}}$. For $n < k \leqslant m$, we have
$$\frac{\pi}{4}= \frac{1}{(2n)^2 x_N^2} \leqslant \frac{1}{(x_Nk)^2} < \frac{1}{\lceil \sqrt{2}N\rceil^2 x_N^2}< \frac{\pi}{2}, $$
and
$$\frac{1}{\sqrt{2}} < \sin \frac{1}{(x_N k)^2} < 1$$
Hence,
$$\left|\sum_{k=n+1}^{m}x_N\sin \frac{1}{(x_Nk)^2} \right| = \sum_{k=n+1}^{m}x_N\sin \frac{1}{(x_Nk)^2}> (2N - \lceil{\sqrt{2}N\rceil})\cdot \frac{1}{N \sqrt{\pi}}\cdot \frac{1}{\sqrt{2}} \\ > \frac{2 - \sqrt{2}}{\sqrt{2 \pi}},$$
which violates the Cauchy criterion for uniform convergence (when $\epsilon$ less than the RHS is chosen).
