<s>[INST] Convergence of $\sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2}$. My try. Examine the convergence of:
$$\sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2}$$

Pointwise convergence (for every $ x \in \mathbb{R}$ ):
$\displaystyle \lim_{n \to \infty} x\frac{\sin(n^2x)}{n^2} = 0$ , because $\sin(n^2x) \in [-1, 1]$

Uniform convergence (for every $ x \in \mathbb{R}$ and $n \in \mathbb{N}$ ):
form dirichlet criterion:
$$\displaystyle \sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2} = \displaystyle \sum_{n=1}^{\infty} f_n(x) \cdot g_n(x) $$ where:

*

*$f_n(x) = \frac{\sin(n^2x)}{n} \leq 2$ which works for: $ \forall x \in \mathbb{R}, n \in \mathbb{N}$

*$g_n(x) = \frac{x}{n} \implies \displaystyle \lim_{n \to \infty} g_n(x) = 0$ (monotonously)

Therefore the series is uniformly convergent for $x \in \mathbb{R}$.
Is that correct?
 [/INST] In the comments you can see the issues with your solution. Here I present a possible approach to this problem. The pointwise convergence follows from the estimate
$$
\left| {x\sum\limits_{n = 1}^\infty  {\frac{{\sin (n^2 x)}}{{n^2 }}} } \right| \le \left| x \right|\sum\limits_{n = 1}^\infty  {\frac{{\left| {\sin (n^2 x)} \right|}}{{n^2 }}}  \le \left| x \right|\sum\limits_{n = 1}^\infty  {\frac{1}{{n^2 }}}  = \frac{{\pi ^2 }}{6}\left| x \right|.
$$
Note that
\begin{align*}
\mathop {\sup }\limits_{x \in \mathbb{R}} \left| {x\sum\limits_{n = 2N}^\infty  {\frac{{\sin (n^2 x)}}{{n^2 }}} } \right| & \ge \left| {\frac{\pi }{2}(2N + 1)\sum\limits_{n = 2N}^\infty  {\frac{{\sin \left( {n^2 \frac{\pi }{2}(2N + 1)} \right)}}{{n^2 }}} } \right|
\\ &
 = \frac{\pi }{2}(2N + 1)\sum\limits_{k = 0}^\infty  {\frac{1}{{(2N + 2k + 1)^2 }}} \\ &  \ge \frac{\pi }{2}(2N + 1)\sum\limits_{k = 0}^\infty  {\frac{1}{{(2N + 2k + 1)(2N + 2k + 3)}}} 
\\ &
 = \frac{\pi }{4}(2N + 1)\sum\limits_{k = 0}^\infty  {\left( {\frac{1}{{2N + 2k + 1}} - \frac{1}{{2N + 2k + 3}}} \right)}  = \frac{\pi }{4}
\end{align*}
for any positive integer $N$. Thus the series cannot be uniformly convergent.
</s>