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 $|S_{n+p}-S_n|\lt \varepsilon$?

Note that
$$3^n \sin\left(\frac{1}{4^nx}\right)\sim \frac{3^n}{4^nx}$$
and by ratio test $\sum \frac{3^n}{4^nx}$ converges $\forall x\neq 0$, therefore the given series converges by limit comparison test with $\sum \frac{3^n}{4^nx}$.
