$\sum_{n=1}^{\infty} {2^n}{\sin(\frac{1}{3^nx})}$ is uniformly continuous on the Interval $(1,\infty)$? Is $\sum_{n=1}^{\infty} {2^n}{\sin(\frac{1}{3^nx})}$ uniformly continuous on the Interval $(1,\infty)$ ?
I have tried:
$$M_{n} = \max|{2^n}{\sin(\frac{1}{3^nx})}| < \max|{2^n}{\frac{1}{3^nx}}|  = \max (\frac{2}{3})^n \frac{1}{x}  < (\frac{2}{3})^n$$
So this should be uniformly convergent. 
I don't know the correct answer. 
Have I missed something here? This should be uniformly convergent correct?
 A: You have proved that the series is uniformly convergent. This proves that the sum is continous. To prove that it is uniformly continuous on $[0,\infty)$ observe that $|\sum\limits_{n=1}^{\infty} 2^{n} \sin (\frac 1 {3^{n}x})| <\sum\limits_{n=1}^{\infty} 2^{n}  \frac 1 {3^{n}x} <\epsilon$ if  $x >\frac 2 {\epsilon}$. If a function is continuous on $[0,\infty)$ and vanishes at $\infty$ then it is uniformly continuous. To prove this let $f$ be the sum of the series and  $\epsilon >0$ and choose $M$ such that $|f(x)| <\epsilon /2$ for $x >M$.(Note that the series actually converges uniformly on $[1,\infty)$ there is no problem near $1$).  Since $f$ is continuous on the compact interval $[1 ,M+1]$ there exits $\delta \in (0,1)$ such that  $x,y \in [1  ,M+1],|x-y| <\delta$ implies $|f(x)-f(y)| <\epsilon$. You can now verify that $x,y \in (1  ,\infty],|x-y| <\delta$ implies $|f(x)-f(y)| <\epsilon$.
A: Yes, we have $|{2^n}{\sin(\frac{1}{3^nx})}| \le (\frac{2}{3})^n$ for all $x > 1$ and all $n$. Since $ \sum (\frac{2}{3})^n$ is convergent, the Weierstraß test shows that the series is uniformly convergent in $(1, \infty).$
