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

Question

I have to show that the series $\sum_{n=1}^{\infty} \frac{1}{n} \sin\bigl( \frac{x^2}{n} \bigr)$ is pointwise convergent for $-\infty<x<\infty$, and that it is uniformly convergent on any interval of the form $[-K,K]$ for $0<K<\infty$.

I can show that the sequence $\{f_n\}$ is pointwise convergent, but I can't seem to link the convergence of the sequence to the convergence of the series. Likewise with the uniform convergence of the series.

I cannot find a valid series for doing the Weierstass' M-test.

Answer 1

Usince the fact that $(\forall x\in\mathbb{R}):\bigl|\sin(x)\bigr|\leqslant|x|$, one gets that$$\sum_{n=1}^\infty\left|\frac1n\sin\left(\frac{x^2}n\right)\right|\leqslant\sum_{n=1}^\infty\frac{x^2}{n^2}.$$Therefore, if $x\in[-K,K]$, one has$$\sum_{n=1}^\infty\left|\frac1n\sin\left(\frac{x^2}n\right)\right|\leqslant\sum_{n=1}^\infty\frac{K^2}{n^2}$$and this proves that the series converges uniformly on $[-K,K]$.