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Let $(c_n)$ be a decreasing sequence of positive numbers. If $\sum c_n \sin nx$ is uniformly convergent, then how to show that $\lim_{n\to\infty} n c_n = 0$?

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This isn't true in general. I think you want to assume you have uniform convergence on an interval of the form $I=[0,\delta]$. –  David Mitra Feb 7 '13 at 16:48
On what basis do you say that? You've not given any counter-examples. –  Saaqib Mahmuud Feb 15 '13 at 19:57
See this. So, for example, $\sum{1\over\sqrt n}\sin(nx)$ converges uniformly on $[\delta,2\pi-\delta]$, $\delta>0$ (but not on an interval containing $0$; hence my previous comment). I'm not sure how to prove your result, with the assumption of uniform convergence on an interval containing 0. –  David Mitra Feb 15 '13 at 20:25

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Hint: If $\sum u_n$ be a convergent series of positive real numbers and $\{u_n\}$ is a monotone decreasing sequence then $\lim_{n\rightarrow\infty} nu_n=0$

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Can you please also fill in the details and then also complete the proof of the original assertion? –  Saaqib Mahmuud Feb 7 '13 at 20:41
I'm still looking for a conclusive proof of my statement. Can anyone enlighten me on how to do it? –  Saaqib Mahmuud Feb 15 '13 at 19:58
While this hint is not directly applicable (for example, $a_n=\frac{1}{n\log n}$ doesn't converge in sum), the proof of her hint using Cauchy criterion can be adapted to prove your result. Here is the proof of her hint, see if you can adapt it to your problem: math.stackexchange.com/a/369824/29892 this one is from @robjohn. There are of course other proofs of her hint (one nice proof combines Cauchy condensation test and the squeeze theorem), but the one I linked to is more easily adaptable to your problem. –  Marcel T. yesterday

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