Uniform convergence of series: $\sum_{n=1}^{\infty}{2^n\sin\left(\frac{x}{3^n}\right)}$

Question

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

Answer 1

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!

Answer 2

Hint:

If $t>0$ then $\sin t<t$.

Answer 3

$$|\sin{\frac{x}{3^n}}| \leq \frac{|x|}{3^n}$$ 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.