Convergence of $\sum_{n=1}^{\infty} 3^n \sin(\frac{1}{4^nx})$

Question

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

Answer 1

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 $\sum \frac{3^n}{4^nx}$.

Answer 2

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, $$|S_{n+p}-S_n|\le 3^{n+1}\dfrac1{4^{n+1}}+\dots3^{n+p}\dfrac1{4^{n+p}}=\frac{\bigl(\frac 34\bigr)^{n+1}-\bigl(\frac 34\bigr)^{n+p+1}}{\frac14}<4\,\Bigl(\frac 34\Bigr)^{n+1}.$$