Is $ S_n$ is uniformly convergent ? Yes/NO 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

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.
