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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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    $\frac{1}{n}\le 1$, for $n>0$. Am I missing something? If you are thinking to $1/0=\infty$, then don't. – 2017-01-03
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    I believe that you're confusing the _sequence_ $\frac{1}{n}$, which is bounded, with the _series_ $\sum \frac{1}{n}$ which is not. – 2017-01-03

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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.

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The sequence is bounded from above by $9$. For every $n\in \mathbb N$, the inequality $$\frac1n\leq 9$$ is true.

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    how about 1/n where n=0.1? – 2017-01-03
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    @LoveScience In a **sequence**, the terms only come from $\mathbb N$. So, you can speak of $a_1$, $a_2$, $a_{84923}$, but there is no such thing as $a_{0.1}$. – 2017-01-03
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    @LoveScience YOU'RE WELCOME!!!!!!!!!!!!!!! :) – 2017-01-03
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    We recommend you avoid all caps. Thank you and have a nice day – 2017-02-04
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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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    but for say n =0.1 the sequence gives you 10 which is bigger than 2 – 2017-01-03
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    @LoveScience, do you know what sequence is? – 2017-01-03
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    @LoveScience, a sequence only uses $n = 1,2,3,\ldots$. – 2017-01-03