Absolutely convergent but not uniformly convergent

[samithie]

I'm trying to show \sum_{k=1}^{\infty}2^{k}sin(\frac{1}{3^{k}x}) does not converge uniformly on any (epsilon, infinity)

now I was able to show that it converges absolutely for x nonzero, by getting it in the form \sum_{k=1}^{\infty}x\left(\frac{2}{3}\right)^{k}\frac{sinx}{x} 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, \sum_{k=n}^{m}2^{k}sin(\frac{1}{3^{k}x}) needs to be less than any given \varepsilon

 

[Stephen Tashi]

can't get tex to work

You must use XML-like "tags" before and after the LaTex. The tags are "itex" and "/itex" enclosed inside square brackets for inline LaTex and "tex" and "/tex" for LaTex on its own line.

I'm trying to show \sum_{k=1}^{\infty}2^{k}sin(\frac{1}{3^{k}x}) does not converge uniformly on any (\epsilon, \infty)