prove that $\lim_{}\frac{1}{3^{n}} = 0$ Because
$$\frac{1}{3^{n}}< \frac{1}{3^{n-2}}$$
$$\frac{1}{3^{n-2}}< \epsilon $$
$$\frac{1}{\epsilon }< 3^{n-2} $$
and
$$\frac{1}{3^{n-2} }<\epsilon$$
because
$$ \frac{1}{3^{n}}< \frac{1}{3^{n-2}} and \frac{1}{3^{n-2}}< \epsilon $$
By transitivity property
$$\frac{1}{3^{n}}< \epsilon $$
therefore the limit converge to 0
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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\begin{align}
&\forall\ \epsilon > 0\quad \mbox{let}\quad
N = \left\lfloor-\,{\ln\pars{\epsilon} \over \ln\pars{3}}\right\rfloor.\qquad
\mbox{Then,}\quad
\forall\ \epsilon > 0\,,\quad n > N \implies {1 \over 3^{n}} <\epsilon
\\[5mm] & \implies
\bbx{\ds{\lim_{n \to \infty}{1 \over 3^{n}} = 0}}
\end{align}
A: 3^n is unbounded and increasing. Thus 1/3^n gets arbitrarily close to 0 and is decreasing. This is sufficient.
