$\sum_{n=1}^{\infty} {2^n}{\sin(\frac{1}{3^nx})}$ 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:

$$M_{n} = \max|{2^n}{\sin(\frac{1}{3^nx})}| &lt; \max|{2^n}{\frac{1}{3^nx}}|  = \max (\frac{2}{3})^n \frac{1}{x}  &lt; (\frac{2}{3})^n$$

So this should be uniformly convergent. 

I don't know the correct answer. 
Have I missed something here? This should be uniformly convergent correct?
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 $|\sum\limits_{n=1}^{\infty} 2^{n} \sin (\frac 1 {3^{n}x})| &lt;\sum\limits_{n=1}^{\infty} 2^{n}  \frac 1 {3^{n}x} &lt;\epsilon$ if  $x &gt;\frac 2 {\epsilon}$. 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  $\epsilon &gt;0$ and choose $M$ such that $|f(x)| &lt;\epsilon /2$ for $x &gt;M$.(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  $x,y \in [1  ,M+1],|x-y| &lt;\delta$ implies $|f(x)-f(y)| &lt;\epsilon$. You can now verify that $x,y \in (1  ,\infty],|x-y| &lt;\delta$ implies $|f(x)-f(y)| &lt;\epsilon$.
Yes, we have $|{2^n}{\sin(\frac{1}{3^nx})}| \le (\frac{2}{3})^n$ for all $x &gt; 1$ and all $n$. Since $ \sum (\frac{2}{3})^n$ is convergent, the Weierstraß test shows that the series is uniformly convergent in $(1, \infty).$