# Prove that $\sum_{n=1}^\infty \frac{1}{n} \sin(\frac{x^2}{n})$ is pointwise convergent for $x \in \mathbb{R}$

For $$x \in \mathbb{R}$$ consider the series $$S = \sum_{n=1}^\infty \frac{1}{n} \sin(\frac{x^2}{n})$$ Then I have to prove that $$S$$ converges pointwise.

My attempt:

It follows from the mean value theorem that $$|\sin(\frac{x^2}{n})| \leq \frac{|x^2|}{n}$$ Thus for $$x \in [-K,K]$$ where $$0 < K < \infty$$ we have that $$\left| \frac{1}{n} \sin(\frac{x^2}{n}) \right| \leq \frac{1}{n} \frac{|x^2|}{n} \leq \frac{K^2}{n^2}$$ where $$K^2 \sum_{n=1}^\infty \frac{1}{n^2}$$ converges pointwise (I am not sure here whether I should just say converges or pointwise converges). Thus it follows form the comparison criteria that $$S$$ converges pointwise (should I then again first say converges and thus also pointwise converges)?

Thanks for your time and help.

• You can get properly sized parentheses (and other paired delimiters) that adjust to the size of their content by preceding them with \left and \right. – joriki May 25 at 11:09
• I suspect you mean the mean value theorem? – joriki May 25 at 11:10
• Oh yes. I am not always sure what the name is in English as my book is in Danish and I am just trying to translate to English. – Mathias May 25 at 11:11
• You mean $\left|\sin\left(\frac{x^2}{n}\right)\right|\le\frac{|x^2|}{n}$. – TonyK May 25 at 11:18
• Oh ye sure. That was a typo! – Mathias May 25 at 11:19

$$\forall t:|\sin t|\le|t|$$ and
$$\left|\sum_{n=1}^\infty \frac1n\sin\frac{x^2}n\right|\le x^2\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2x^2}6.$$
• Thanks. It doesn't matter if $x \in \mathbb{R}$ or if $x \in [-K,K]$ where $0 < K < \infty$? Would it matter if I wanted to use Weiterstrass' M-test? – Mathias May 25 at 11:25
• @Mathias: what is the point introducing that $K$ when you have a straightforward formula ? – Yves Daoust May 25 at 12:09