# uniform convergence of $\sum\limits_{n=1}^{\infty}3^n\sin\left(\frac 1 {4^nx}\right)$ in $[1,\infty)$

I want to check the uniform convergence of $\sum\limits_{n=1}^{\infty}3^n\sin\left(\frac 1 {4^nx}\right)$ in $[1,\infty)$. The hint in the book was using Cauchy condition for uniform convergence. For questions of proving a series doesn't converge uniformally, we can take x values s.t the sum is bigger than epsilon.$\forall\epsilon>0$ it's clear I need to find $N\in\mathbb N$ s.t $\forall n>N,p\in\mathbb N,x\in[1,\infty)$ the sum $$\left|\sum_{k=n+1}^{n+p}3^k\sin\left(\frac 1 {4^kx}\right)\right|<\epsilon.$$ Which $N$ can I choose?

• I had a little typing mistake (-1 instead 1), but it had been fixed. – user65985 Jul 4 '13 at 15:05
• Note that it holds that $|\sin t| < t$ for $t > 0$. I think you can show uniform convergence of $\sum_n \frac{3^n}{4^n x}$ on your interval without too much trouble. – Jakub Konieczny Jul 4 '13 at 15:09

This follows immediately from geometric series and from the fact that $|\sin y|\leq |y|$ for every $y\in\mathbb{R}$. Which can be shown by the MVT, for instance.
Added: so it follows from the Weierstrass M-test. Or directly, from where you stopped with the Cauchy criterion: $$\Big|\sum_{k=n+1}^{n+p}3^k\sin \left(\frac{1}{4^kx}\right)\Big|\leq \sum_{k=n+1}^{n+p}\left(\frac{3}{4}\right)^k=\frac{\left(\frac{3}{4}\right)^{n+1}-\left(\frac{3}{4}\right)^{n+p+1}}{1-\frac{3}{4}}\leq 4\left(\frac{3}{4}\right)^{n+1}$$ for every $|x|\geq 1$. I think you can see how to choose $N$ now.
• and then it follows by wirestrass M test with $(\frac 3 4)^n$, right? thank you. – user65985 Jul 4 '13 at 15:15
• @CoarguAliquis It follows directly from your argument. (This is what's going on in the $M$-test...) – David Mitra Jul 4 '13 at 15:16