I need to show that $\sum_{n=1}^\infty \frac{\sin(n^4 x)}{n^2}$ is continuous on $\mathbb{R}$.

'Proof': Let $f_n(x)=\frac{\sin(n^4x)}{n^2}$. It is clear that $f_n(x)$ is continuous for each $n \in \mathbb{N}$. It is clear also that $$ \left| \sum_{n=1}^\infty \frac{\sin(n^4 x)}{n^2} \right| \leq \sum_{n=1}^\infty \left|\frac{\sin(n^4 x)}{n^2} \right| \leq \sum_{n=1}^\infty \frac{1}{n^2}<\infty $$ so that by the Weierstrass M-test, $\sum_{n=1}^\infty f_n(x)=\sum_{n=1}^\infty \frac{\sin(n^4 x)}{n^2}$ converges uniformly on $\mathbb{R}$. That is, $g_m(x)=\sum_{n=1}^m f_n(x)$ converges uniformly. Each $g_m(x)$ is continuous as it is the finite sum of continuous functions. But then $g_m$ is a sequence of continuous functions that converges uniformly so that the limit, namely $\sum_{n=1}^\infty \frac{\sin(n^4x)}{n^2}$ is continuous.

Does this proof work?

  • 2
    $\begingroup$ Yes that works perfectly! Can't think of anything to criticize. $\endgroup$ – Winther Jan 10 '16 at 5:48
  • $\begingroup$ @Winther To remove this from the unanswered question list, feel free to post this as an answer and I will happily accept it for the effort of checking it! $\endgroup$ – Kyle L Aug 14 '16 at 14:34

Your Answer

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

Browse other questions tagged or ask your own question.