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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.

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  • $\begingroup$ 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. $\endgroup$ – joriki May 25 at 11:09
  • $\begingroup$ I suspect you mean the mean value theorem? $\endgroup$ – joriki May 25 at 11:10
  • $\begingroup$ 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. $\endgroup$ – Mathias May 25 at 11:11
  • $\begingroup$ You mean $\left|\sin\left(\frac{x^2}{n}\right)\right|\le\frac{|x^2|}{n}$. $\endgroup$ – TonyK May 25 at 11:18
  • $\begingroup$ Oh ye sure. That was a typo! $\endgroup$ – Mathias May 25 at 11:19
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$$\forall t:|\sin t|\le|t|$$ and

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

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  • $\begingroup$ 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? $\endgroup$ – Mathias May 25 at 11:25
  • $\begingroup$ @Mathias: what is the point introducing that $K$ when you have a straightforward formula ? $\endgroup$ – Yves Daoust May 25 at 12:09
  • $\begingroup$ Weistrass M-test will be useful for uniform convergence on a compact. $\endgroup$ – EDX May 25 at 12:10
  • $\begingroup$ I am not sure. It is an exam question in my Analysis course where the professor has given it as a hint. $\endgroup$ – Mathias May 25 at 12:10
  • $\begingroup$ The question is about pointwise convergence, Weierstrass is overkill. $\endgroup$ – Yves Daoust May 25 at 12:10

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