<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] If the series converged uniformly on $\mathbb{R}$, then for any $\epsilon > 0$ there exists $2N+1 \in \mathbb{N}$ such that for all $x \in \mathbb{R}$,
$$\left|\sum_{n= 2N+1}^\infty x \frac{\sin n^2 x}{n^2} \right| < \epsilon$$
In particular, for $x_m = \frac{\pi}{2} + 2m\pi$ we have $\sin n^2x_m = \sin \left( n^2\frac{\pi}{2}+ 2mn^2 \pi\right)= \sin n^2\frac{\pi}{2}$ and
$$\left|\sum_{n= 2N+1}^\infty  \frac{\sin n^2 \frac{\pi}{2}}{n^2} \right| < \frac{\epsilon}{\frac{\pi}{2} + 2m\pi}$$
Since the RHS converges to $0$ as $m \to \infty$ and the LHS is independent of $m$, we must have
$$\tag{*}\sum_{n= 2N+1}^\infty  \frac{\sin n^2 \frac{\pi}{2}}{n^2}=0$$
However, we can write
$$\sum_{n= 2N+1}^\infty  \frac{\sin n^2 \frac{\pi}{2}}{n^2}= \sum_{j=0}^\infty  \frac{\sin (2N+1+j)^2 \frac{\pi}{2}}{(2N+1+j)^2}$$
where $\sin (2N+1+j)^2 \frac{\pi}{2} = 1$ if $j$ is even and  $\sin (2N+1+j)^2 \frac{\pi}{2} = 0$ if $j$ is odd.
This implies
$$\sum_{n= 2N+1}^\infty  \frac{\sin n^2 \frac{\pi}{2}}{n^2}= \sum_{k=0}^\infty  \frac{1}{(2N+1+2k)^2} \neq 0,$$
which contradicts (*).
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