previously I have proved that the following series converges uniformly in $[a,\infty) ,a>0$ $$ \sum\limits_{n=1}^\infty2^n\sin(\frac{1}{3^nx})$$

But I was requested to prove that it doesn't converge uniformly in $(0,\infty)$ Any hint on what theory could be helpful here?

  • $\begingroup$ Have you tried to use $\sin(X)\ge \frac{2}{\pi}X$ $\endgroup$ – hamam_Abdallah May 30 at 21:55

If the series converged uniformly on $(0, +\infty)$, then each term would also converge uniformly to zero on the same interval (that's a consequence of the Cauchy criterion). And it's fairly easy to show that it's not the case: For $x\in [0, \frac \pi 2]$, we have $\sin x \geq \frac 2 \pi x$. Thus, if $\frac {2}{3^{n}\pi}<x <\frac {2}{3^{n-1}\pi}$

$$2^n\sin(\frac{1}{3^nx}) \geq \frac {2^{n+1}} {3^n \pi} \frac 1 x\geq \frac {2^n} 3$$

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