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.
$$\forall t:|\sin t|\le|t|$$ and
$$\left|\sum_{n=1}^\infty \frac1n\sin\frac{x^2}n\right|\le x^2\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2x^2}6.$$