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The series is $$\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^2}x^3.$$

We know that every $f_n(x)$ is continuous in $\Bbb R$. I wanted to apply some methods that need the series converges uniformly: Weierstrass M-test, if each $f_n(x)$ is continuous and the series converges uniformly then the series is also continuous, Cauchy's test for the uniform convergence to the sequence of partial sums but the problem is that I can't (or don't know) bound any of the $|f_n(x)|$ or $|S_r(x)−S_k(x)|$, where $S_n(x)=\sum_{j=1}^{n}\frac{\sin(jx)}{j^2}x^3$.

So, those are some basic methods I used for this exercise. Any help?

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  • $\begingroup$ just observe that $|\sin(nx)|\le 1$ $\endgroup$
    – Masacroso
    Jul 28, 2017 at 12:45
  • $\begingroup$ You need to bound $|S_{\infty}(x)-S_k(x)| = |x^3\sum_{n=k+1}^\infty \frac{\sin(nx)}{n^2}|$ (easy) $\endgroup$
    – reuns
    Jul 28, 2017 at 12:47
  • $\begingroup$ But the problem of bounding is because the $x^3$, I can't in the interval $\Bbb R$. $\endgroup$ Jul 28, 2017 at 13:24

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Consider an interval $[a,b]$, $|x^3<A$ on this interval and $|f_n(x)|\leq {A\over n^2}$ on this interval. This implies that $f_n$ converges uniformly on $[a,b]$ so is continuous.

https://en.wikipedia.org/wiki/Normal_convergence

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  • $\begingroup$ Yes, that's right in that interval. I mean the convergence in $\Bbb R$. $\endgroup$ Jul 28, 2017 at 13:22
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    $\begingroup$ But if it is continuous on every closed interval it is continuous on $R$, continuity is a local property. $\endgroup$ Jul 28, 2017 at 13:23
  • $\begingroup$ Okey, now I see! And the uniformly convergence of that functional series in $\Bbb R$ is true or not? $\endgroup$ Jul 28, 2017 at 15:04

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