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Suppose that the series $\sum_{1}^{\infty}n|b_n|$ converges. Show that the series $\sum_{1}^{\infty}b_n \sin{nx}$ converges uniformly on $\mathbb{R}$, and that it can be integrated and differentiated term by term.

So I have to use the Weierstrass M-test. Put $M_n=\sup_{x \in \mathbb{R}}|b_n \sin{nx}|=|b_n|$. Now we want to show that $\sum_{1}^{\infty}M_n =\sum_{1}^{\infty}|b_n|$converges then $\sum_{1}^{\infty}b_n \sin{nx}$ converges uniformly.

Do I need to bound $\sum_{1}^{\infty}M_n$ by $\sum_{1}^{\infty}n|b_n|$? I understand the that it can be integrated and differentiated term by term if $\sum_{1}^{\infty}b_n \sin{nx}$ converges uniformly on $\mathbb{R}$.

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  • $\begingroup$ Weierstrass M-Test is just another way of saying the Absolute Convergence Test, except it applies to complex sequences. Im really not sure what the difference is, and if there isnt a practical one then it seems rather trivial to solve. $\endgroup$ – CogitoErgoCogitoSum Dec 5 '14 at 1:05
  • $\begingroup$ I understand that but how do i show that the series converges uniformly @CogitoErgoCogitoSum $\endgroup$ – Black Dexter Dec 5 '14 at 1:07
  • $\begingroup$ Isn't this a comparison of $b_n \leq nb_n$? $\endgroup$ – IAmNoOne Dec 5 '14 at 1:07
  • $\begingroup$ $b_n \sin(nx) \le b_n \le |b_n| \le n |b_n|$ $\endgroup$ – CogitoErgoCogitoSum Dec 5 '14 at 1:10
  • $\begingroup$ Got it @CogitoErgoCogitoSum Thanks a lot $\endgroup$ – Black Dexter Dec 5 '14 at 1:11
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The series $\sum b_n\sin nx$ converges uniformly by Weierstrass M-test; here it is enough to estimate $|b_n\sin nx|\leqslant |b_n|$ and use that $$ \sum |b_n|\leqslant \sum n|b_n|<\infty . $$ A similar argument holds for the differentiated series $\sum nb_n\cos nx$ and the integrated one. Note that termwise differentiation and integration is automatic for power series but not for general series.

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