Uniform convergence of series: $\sum_{n=1}^{\infty}{2^n\sin\left(\frac{x}{3^n}\right)}$ Task: we should find area of x>=0, that the series is Uniform convergence on this area
$$\sum_{n=1}^{\infty}{2^n\sin\left(\frac{x}{3^n}\right)}$$
There are some ways to proof "Uniform convergence of sum". 
I tried to use Dirichlet and Abel's ways to solve, but they are not suited on this case.
However, I checked answers and got: it is not converges uniformly
 A: $$|\sin{\frac{x}{3^n}}| \leq \frac{|x|}{3^n}$$ so the series converges pointwise on the real line.
The series also converges uniformly in every bounded subset of $\Bbb{R}$ by the $M-$test of Weierstrass.
A: Hint:
If $t>0$ then $\sin t<t$.
A: By using the definition, for a bounded subset $X$ of $\mathbb{R}$, you can check whether
$$\lim_{N \to +\infty} \sup_{x \in X} \left|\sum_{n = N+1}^{+\infty} 2^n \sin\left(\frac{x}{3^n}\right)\right| = 0.$$
Using triangular inequality, you get:
$$\left|\sum_{n = N+1}^{+\infty} 2^n \sin\left(\frac{x}{3^n}\right)\right| \leq \sum_{n = N+1}^{+\infty} \left|2^n \sin\left(\frac{x}{3^n}\right)\right|.$$
Since $|\sin(y)| \leq |y|$ for any $y \in \mathbb{R}$, then:
$$\sum_{n = N+1}^{+\infty} \left|2^n \sin\left(\frac{x}{3^n}\right)\right| \leq \sum_{n = N+1}^{+\infty}\left|x \frac{2^n}{3^n}\right| = |x| \sum_{n=N+1}^{+\infty}\left(\frac{2}{3}\right)^n .$$
Using the formula for a geometric series, you get:
$$\sum_{n = N+1}^{+\infty} \left|2^n \sin\left(\frac{x}{3^n}\right)\right| \leq |x|3 \left(\frac{2}{3}\right)^{N+1}.$$
Now, observe that:
$$\lim_{N \to +\infty}\sup_{x \in X} \left|\sum_{n = N+1}^{+\infty} 2^n \sin\left(\frac{x}{3^n}\right)\right| \leq \lim_{N \to +\infty}\sup_{x \in X} |x|3 \left(\frac{2}{3}\right)^{N+1} = 0,$$
and you are done!
