Consider a series $\sum_{n=1}^\infty f_n(x)$, where $f_n(x) = x \sin\frac{1}{x^2n^2}$ and $x \in (0, +\infty)$. If one fix an arbitrary $x \in (0, +\infty)$, then for sufficiently large $n \in \mathbb{N}$ one would have $$ |f_n(x)| = \left|x \sin\frac{1}{x^2 n^2}\right| < x\frac{1}{x^2 n^2}. $$ The series $\sum_{n=1}^\infty \frac{1}{xn^2}$ converges for any fixed $x \in (0, +\infty)$, so $\sum_{n=1}^\infty f_n(x)$ converges pointwise for $x \in (0, +\infty)$. But does $\sum_{n=1}^\infty f_n(x)$ converge uniformly on $(0, +\infty)$?
It is clear, that if $x \in [1, +\infty)$, then $0 < \sin\frac{1}{x^2n^2} < \frac{1}{x^2n^2}, \; \forall n \in \mathbb{N}, $ $$ \left|x \sin\frac{1}{x^2 n^2}\right| < \frac{1}{xn^2} < \frac{1}{n^2}, $$ so $\sum_{n=1}^\infty x \sin\frac{1}{x^2 n^2}$ converges uniformly on $[1, +\infty)$. But I don't understand what we can do with the series when $x \in (0,1)$.
Any help would be appreciated.
If the series converges uniformly on $(0,1)$, then for every $\epsilon > 0$ there exists a positive integer $N$ such that for all $m > n > N$ and $x \in (0,1)$ we have $$\tag{1}|S_{n,m}(x)|:=\left|\sum_{k=n+1}^{m}x \sin \frac{1}{k^2 x^2}\right|=\left|\sum_{k=n+1}^{m}\frac{1}{k^2x} \frac{\sin \frac{1}{k^2 x^2}}{\frac{1}{k^2 x^2}}\right|< \epsilon$$
Taking $m = 2n$ and $x = n^{-1}\in (0,1)$ we get
$$\tag{2}|S_{n,2n}(n^{-1})|=\left|\sum_{k=n+1}^{2n}\frac{n}{k^2} \frac{\sin \frac{n^2}{k^2 }}{\frac{n^2}{k^2 }}\right|$$
Since $\frac{n^2}{k^2} <1$ for $k \geqslant n+1$ it follows that $\frac{\sin \frac{n^2}{k^2 }}{\frac{n^2}{k^2 }} >\sin(1)>0$ and (removing the absolute value symbol on the RHS of (2))
$$|S_{n,2n}(n^{-1})|= \sum_{k=n+1}^{2n}\frac{n}{k^2} \frac{\sin \frac{n^2}{k^2 }}{\frac{n^2}{k^2 }}> n \cdot \frac{n}{(2n)^2}\cdot \sin(1)= \frac{\sin(1)}{4}$$
Hence, condition (1) is violated for any $\epsilon < \frac{\sin(1)}{4}$ no matter how large we take $n$. Consequently, the series does not converge uniformly for $x \in (0,1)$.