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?

  • $\begingroup$ 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

1 Answer 1


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.


  • $\begingroup$ Yes, that's right in that interval. I mean the convergence in $\Bbb R$. $\endgroup$ Jul 28, 2017 at 13:22
  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.