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Is $\displaystyle S_n = \sum_{n=1}^{\infty} 2^n \sin\frac{1}{3^nx}$ uniformly convergent on the interval $[1, \infty)$ ? True /false

My attempt : NO, the given series will not uniformly convergent on $[1, \infty)$ because the cauchy sequence criterion for uniform convergence fail

That is $$\vert S_{n +m}(x) - S_n(x) |= 2^{n+1} \sin \frac{1}{3^{n+1}x} + ....... + 2^{n+m} \sin \frac{1}{3^{n+m}x} \ge 2^{n+1} \frac{2}{ \pi} \frac{1}{3^{n+1} x} +....... + 2^{n+m}\frac{2}{\pi} \frac{1}{3^{n+m}x} \ge \frac{2^{n+1} {2}}{\pi3^{n+1} x}$$

Is this true ?

Any hints/solution will be appreciated

thanks u

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    $\begingroup$ Your calculation looks correct, however the conclusion you draw is wrong. You show that $|S_{n+m}-S_n|$ is greater than something that goes to zero when $n\to\infty$. This tells you nothing about if the Cauchy criterion fails or not. $\endgroup$ – Winther May 14 '19 at 23:03
  • $\begingroup$ @Winther if put $x= 1/ 3^n$ then its fail $\endgroup$ – jasmine May 14 '19 at 23:06
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    $\begingroup$ But $x=1/3^n$ is not in $[1,\infty)$ right? $\endgroup$ – Winther May 14 '19 at 23:13
  • $\begingroup$ Ya u r right @Winther thanks gots it now $\endgroup$ – jasmine May 14 '19 at 23:14
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True. The main idea is that $\sin x \leq x$ for $x\geq 0.$ Hence, on the domain $[1,\infty),$

$$ \sum_{n=1}^\infty 2^n \sin\left(\frac{1}{3^n x}\right) \leq \sum_{n=1}^\infty 2^n \left(\frac{1}{3^n x}\right) \leq \sum_{n=1}^\infty \left(\frac{2}{3}\right)^n, $$ which is obviously convergent as a geometric series.

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  • $\begingroup$ thanks @ZuhZug but im not getting how u directly writes $\sum_{n=1}^\infty 2^n \left(\frac{1}{3^n x}\right) \leq \sum_{n=1}^\infty \left(\frac{2}{3}\right)^n?$ $\endgroup$ – jasmine May 14 '19 at 23:05
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    $\begingroup$ Fix $2^n.$ The fraction $\frac{1}{3^n x}$ is maximized when $x=1.$ This is true for all $n$ so the inequality holds for the sum. $\endgroup$ – zugzug May 14 '19 at 23:09

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