# Prove that the following is not uniformly convergent

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?

• Have you tried to use $\sin(X)\ge \frac{2}{\pi}X$ – 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 $$\frac {2}{3^{n}\pi}
$$2^n\sin(\frac{1}{3^nx}) \geq \frac {2^{n+1}} {3^n \pi} \frac 1 x\geq \frac {2^n} 3$$