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Can someone please verify my answers?

Consider the series $$\sum\limits_{n=1}^\infty \sin \left(\frac{x}{n^2}\right)$$

Prove that the series converges uniformly on the bounded interval $[-M, M]$, for any $M > 0$.

Note that for all real numbers $y$, $$|\sin y| \leq |y|$$

Therefore, for all $M > 0$, we have$$\left|\sin \left( \frac{x}{n^2}\right)\right| \leq \left|\frac{x}{n^2} \right| \leq \frac{M}{n^2}$$

By the Weierstrass M-test, the series converges uniformly on $[-M, M]$.

To what function does this series converge pointwise?

I'm stuck over here.

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  • 2
    $\begingroup$ That is correct. Regarding the function, I think you have to do a clever substitution and then verify if you recognize a Fourier series $\endgroup$ – AnalysisStudent0414 Jun 13 '14 at 13:15
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To find the limit function, you can just consider the Taylor series of $\sin\left(\frac{x}{n^2}\right)$ and sum over $n$ (the sum switch is legit under uniform convergence): $$\sin z = \sum_{m=0}^{+\infty}\frac{(-1)^m z^{2m+1}}{(2m+1)!},$$ $$\sum_{n=1}^{+\infty}\sin\frac{x}{n^2}=\sum_{m=0}^{+\infty}\frac{(-1)^m \zeta(4m+2)}{(2m+1)!}x^{2m+1}.\tag{1}$$

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