<s>[INST] Convergence of $\sum_{n=1}^{\infty} 3^n \sin(\frac{1}{4^nx})$ I wish to prove the convergence of:
$$\sum_{n=1}^\infty 3^n \sin\left(\frac 1 {4^nx}\right)$$
for $1\le x \lt \infty$, using Cauchy's criterion. 
Here is what I tried:
\begin{align}
|S_{n+p}-S_n| & = \left| 3^{n+1} \sin\left(\frac 1 {4^{n+1}x}\right) 
 + \cdots+3^{n+p} \sin\left(\frac 1 {4^{n+p}x}\right)\right| \\[10pt]
& \le \left|4^{n+1} \sin\left(\frac 1 {4^{n+1}x}\right) \cdots 4^{n+p} \sin\left(\frac 1 {4^{n+p}x}\right) \sin\left(\frac 1 {4^{n+1}x}\right)(4^{n+1}+\cdots+4^{n+p}) \right|
\end{align}
I tried using geometric series sum from here but came empty handed.
how can I show that $|S_{n+p}-S_n|\lt \varepsilon$?
 [/INST] If you absolutely want to use Cauchy's criterion, you can onserve that, if $n$ is large enough, $0<\frac1{4^n x}<\frac\pi2$, hence 
$$0<\sin\frac1{4^n x}<\sin\frac1{4^{n+1}x}<\dotsm<\sin\frac1{4^{n+p}x},$$
and $\;\sin \dfrac1{4^{n+k}x}< \dfrac1{4^{n+k}x}<\dfrac1{4^{n+k}}$, so that, by the  triangle inequality,
$$|S_{n+p}-S_n|\le 3^{n+1}\dfrac1{4^{n+1}}+\dots3^{n+p}\dfrac1{4^{n+p}}=\frac{\bigl(\frac 34\bigr)^{n+1}-\bigl(\frac 34\bigr)^{n+p+1}}{\frac14}<4\,\Bigl(\frac 34\Bigr)^{n+1}.$$
</s>