# Is $S_n$ is uniformly convergent ? Yes/NO

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

• Your calculation looks correct, however the conclusion you draw is wrong. You show that $|S_{n+m}-S_n|$ is greater than something that goes to zero when $n\to\infty$. This tells you nothing about if the Cauchy criterion fails or not. – Winther May 14 '19 at 23:03
• @Winther if put $x= 1/ 3^n$ then its fail – jasmine May 14 '19 at 23:06
• But $x=1/3^n$ is not in $[1,\infty)$ right? – Winther May 14 '19 at 23:13
• Ya u r right @Winther thanks gots it now – jasmine May 14 '19 at 23:14

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.
• thanks @ZuhZug but im not getting how u directly writes $\sum_{n=1}^\infty 2^n \left(\frac{1}{3^n x}\right) \leq \sum_{n=1}^\infty \left(\frac{2}{3}\right)^n?$ – jasmine May 14 '19 at 23:05
• Fix $2^n.$ The fraction $\frac{1}{3^n x}$ is maximized when $x=1.$ This is true for all $n$ so the inequality holds for the sum. – zugzug May 14 '19 at 23:09