I'm trying to show [itex]\sum_{k=1}^{\infty}2^{k}sin(\frac{1}{3^{k}x})[/itex] does not converge uniformly on any (epsilon, infinity)(adsbygoogle = window.adsbygoogle || []).push({});

now I was able to show that it converges absolutely for x nonzero, by getting it in the form [itex]\sum_{k=1}^{\infty}x\left(\frac{2}{3}\right)^{k}\frac{sinx}{x}[/itex] and so the sinx/x < 1 drops out and so its a convergent geometric series. This x is not really x, I just mean its of the form sinx/x. so the series converges to 3x. However I'm not sure about the uniform part, I'm trying to show for any episilon, I can't bound the tail of the series but I don't know that it converges to 3x do I, all I know is that it converges to something smaller than 3x

so I need to bound the tail using cauchy criterion but can't figure out why it's true. ie for n sufficiently large, [itex]\sum_{k=n}^{m}2^{k}sin(\frac{1}{3^{k}x})[/itex] needs to be less than any given [itex]\varepsilon[/itex]

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# Absolutely convergent but not uniformly convergent

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