Showing that the given sequence is bounded as follows Let $A = (x_n)$ be a sequence that defined as $x_n = \frac{1}{3^{n+5}}$.
Show that $A$ is bounded and find it's supremum and the infimum.
Attempt:
First, I claim that $A$ is a decreasing sequence. I show this by induction as follows:
To show $x_{n+1}<x_n$ for all $n \in \Bbb Z^+$. Indeed, this is true for $n=1$. Now, assume that for $n=k$, it's also true; that is s, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Then,
$x_{k+2} = \frac{1}{3^{(k+2)+5}} = \frac{1}{3^{k+7}} < \frac{1}{3^{k+6}} = \frac{1}{3^{(k+1)+5}} = x_{k+1}$. Hence, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Therefore, $x_{n+1} < x_n$
for all $n \in \Bbb Z^+$. Thus, $A$ is a decreasing sequence. $\Box$
Back to the problem. It's clear that $A$ is bounded above by $\frac{1}{3^6}$ (Should I show this first?).
Then, to show that $A$ is bounded, it's suffices to show that $A$ is bounded below by $0$.
I show this one again by induction. Indeed, it's true for $n=1$.
Assume that it's true for $n=k$; that is $0 < x_k$. Then, $x_{k+1} = \frac{1}{3^{(k+1)+5}}
= \frac{1}{3^{k+5}} \cdot \frac{1}{3} > 0$. Hence, $0< x_n$ for all $n \in \Bbb Z^+$. Therefore, $A$ is bounded below by $0$. Thus, $A$ is bounded, as desired.
Now, I claim that $\sup A = \frac{1}{3^6}$ and $\inf(A) = 0$.
For the proof of infimum, let $m$ be an another lower bound of $A$. To show: $m \le 0$. Suppose
$m > 0$. Then, by the Density Theorem, there exists $r \in \Bbb Q$ such that
$0 < r < m$. Hence, $r \in A$. A contradiction, since $m$ is a lower bound of $A$.
Thus, $m \le 0$ and therefore, $\inf A = 0$.
For the supremum, let $M$ be an another upper bound of $A$. To show: $\frac{1}{3^6} \le M$. Suppose $M < \frac{1}{3^6}$. Then, by the Density Theorem, there exists $s \in \Bbb Q$ such that
$M < s < \frac{1}{3^6}$. Hence, $s \in A$, contradiction with the fact that $M$ is an upper bound of $A$.
Therefore, $\frac{1}{3^6} \le M$ and thus, $\sup A = \frac{1}{3^6}$.
Does those approach true?
 A: 
Then, by the Density Theorem, there exists $r \in \Bbb Q$ such that $0
< r < m$.

This is true, but irrelevant.

Hence, $r \in A$. A contradiction, since $m$ is a lower
bound of $A$.

No, you haven't show that $r \in A,$ so no, you haven't shown a contradiction. You correctly showed that there is a rational number between $0$ and $m$, but you haven't (and you won't be able to) get from "$r$ is a rational number" to "$r$ is a member of $A$".
In summary, you haven't shown that $\ r \in A.$ Think about what it means to show that $\ r \in A $ [Hint: what is $A$?]
You could, however, amend your argument as follows:
For the proof of infimum, let $m$ be an another lower bound of $A$. To show: $m \le 0$. Suppose
$m > 0$. Then, since [$\color{blue}{\text{*insert reasoning here*}}$], there exists $\color{blue}{r \in A}$ such that
$0 < r < m$. Hence, $r \in A$. A contradiction, since $m$ is a lower bound of $A$.
Thus, $m \le 0$ and therefore, $\inf A = 0$.
A: Using the "Archimedean property" , we get for every $\epsilon \gt 0 $ , there exists $m \in \mathbb{N}$ such that $\frac{1}{n} \lt \epsilon$ for every $n\ge m $ .
As, $3^{n+5}\gt n \implies \frac{1}{3^{n+5}} \lt \frac{1}{n} \lt \epsilon $ for all $n\ge m $
By definition of convergence of a sequence, sequence
$\{\frac{1}{3^{n+5}}\}$ converge to $0$.
So, the given sequence is bounded and $limsup=liminf=0$
