Convergence of $\sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2}$. My try.

Question

Examine the convergence of:

$$\sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2}$$

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Pointwise convergence (for every $ x \in \mathbb{R}$ ):

$\displaystyle \lim_{n \to \infty} x\frac{\sin(n^2x)}{n^2} = 0$ , because $\sin(n^2x) \in [-1, 1]$

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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?

Answer 1

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.

Answer 2

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 (*).