Is $ S_n$ is uniformly convergent ? Yes/NO

Question

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

Answer 1

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.