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Given a bounded real sequence $\{a_n\}$, for what values of the real exponent $p$ can you be sure that the series $$\sum_{n=1}^{\infty} \frac{a_n}{n^p} \sin(nx)$$ sums to a twice-continuous differentiable function of $x$? Explain your answer.

Attempt: We note that $|a_n| < M$ where $M>0$. We also know that $\sum_{k=1}^{n}\sin{mx}=\frac{\sin{((n+1)\theta/2)\sin{(n\theta/2)}}}{\sin{(\theta/2)}}$ Thus $|\sum_{k=1}^{n}\sin{nx}|$ is bounded. Now for $\delta >0$ , $n^{-\delta} \downarrow 0$ and thus by Dirichlet, $\sum_{n=1}^{\infty} \frac{a_n}{n^p} \sin(nx)$ converges uniformly.

I am stuck here. How do I find value of $p$ to be twice differentiable?

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  • $\begingroup$ see here: dpmms.cam.ac.uk/~agk22/uniform.pdf care must be taken. You need the $p$ to offset the power of $n$ that will come out in the numerator when you differentiate $\sin(nx)$ $\endgroup$ – qbert Mar 13 '18 at 23:28
  • $\begingroup$ So you would be using Theorem 2 in the document you shared right? And P>2 coz we are differentiating twice? $\endgroup$ – Dom Mar 13 '18 at 23:31

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