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Show that the series $$\sum_{n=1}^{\infty} \frac{x\sin(n^2x)}{n^2}$$ converges pointwise to a continuous function on $\mathbb{R}$

Using the boundedness of $\sin(n^2 x)$ and the convergence of $\sum_{n=1}^{\infty} \frac{1}{n^2}$, it was easy to show pointwise convergence. Showing that the function is continuous is giving me a lot of trouble, however.

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Well, first, note that you can write $$ \sum_{n=1}^{\infty}\frac{x\sin(n^2x)}{n^2}=x\sum_{n=1}^{\infty}\frac{\sin(n^2x)}{n^2}. $$ So, it is sufficient to show that the series on the RHS (without the $x$) converges to a continuous function.

Now, the functions $f_n(x):=\frac{\sin(n^2x)}{n^2}$ are continuous on $\mathbb{R}$; so, to show that the series is continuous, it is sufficient to show that convergence of the series is uniform on $\mathbb{R}$. Can you see how to do that?

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    $\begingroup$ For all $\epsilon > 0$, $\sum_{k=m}^{n} |\frac{\sin (k^2 x)}{k^2}| \leq \sum_{k=m}^{n} \frac{1}{k^2} < \epsilon$ for all $x \in \mathbb{R}$ and $n,m \geq N$ where $N$ is sufficiently large. Is this correct? $\endgroup$ – GillyB Sep 22 '17 at 2:27
  • $\begingroup$ @GillyB Bingo. Nicely done. $\endgroup$ – Xander Henderson Sep 22 '17 at 2:31

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