Since "Every convergent sequence is bounded" and bounded means bounded above and below. Then why is the sequence $\frac 1n$ convergent? since it is not bounded above (converges to $0$ for lim -> infinity) it is bounded below by $0$ but not above so the sequence is not bounded yet it is convergent.
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$\begingroup$There seems to be misunderstanding about definition of sequence:
Sequence $(a_n)$ in set $S$ is a function $a\colon \Bbb N\to S$.
For example, sequence $a_n = \frac 1 n$ is a function $n\mapsto\frac 1 n\colon \Bbb N\to\Bbb R$ and some of its terms are: $$1,\frac 12,\frac 13,\frac 14,\frac 15,\frac 16\ldots$$
You can observe that $a_n> 0$, for all $n\in\Bbb N$:
$$\frac 1 n> 0 \iff n > 0\quad\small\text{(which is true for all $n\in\Bbb N$)}$$ and thus, $(a_n)$ is bounded from bellow.
Also, observe that $(a_n)$ is strictly decreasing sequence, i.e. $a_n> a_{n+1}$, for all $n\in\Bbb N$:
$$\frac 1n>\frac 1{n+1} \iff n+1>n\iff 1>0.$$
In particular, $a_n\leq a_1 = 1$, for all $n\in\Bbb N$ and thus, $(a_n)$ is bounded from above.
$\endgroup$ $\begingroup$The sequence is bounded from above by $9$. For every $n\in \mathbb N$, the inequality $$\frac1n\leq 9$$ is true.
$\endgroup$ 4 $\begingroup$It is bounded above, by, say, 2: every term is less than 2. A bound does not have to be tight.
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