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Since a sequence $\{a_n\}$ is a function whose domain is the set of natural numbers, ie. $a_n$ exists for all $n\in\Bbb{N}$, then $a_1$ always exists for any sequence. Therefore any monotone sequence is bounded by $a_1$: below, if $\{a_n\}$ is increasing; above, if $\{a_n\}$ is decreasing. Moreover, the limit if a sequence is $L$ provided we can make $a_n$ as close to $L$ as possible as $n$ increases to infinity. Then, the sequence $\{a_n\}$ is bounded by $L$: above, if $\{a_n\}$ is increasing; below, if $\{a_n\}$ is decreasing. Therefore, this follows that a sequence is bounded iff it is convergent.

Follow-up questions:

1) Doesn't this imply that a sequence is convergent if it is bounded?

2a) Are periodic sequences with period $1$ legitimate periodic sequences? For example, a sequence defined by $\{a_n\} = \{0, 0, 0, 0, \cdots\}$ for all $n$.

2b) Are sequences bounded below & above by the same number legitimate as well? In the previous example, the sequence $\{a_n\}$, if it is indeed a legitimate sequence, is bounded below and above by $0$.

2c) The Monotone Convergence Theorem (MCT) states that a sequence is convergent if it is monotonous and bounded. From the previous remark, if it is correct, and if the answer to questions 2a) and 2c) are yes, then I don't see why a sequence must be both monotone AND bounded for the MCT to be true. Is it not enough of a condition for the sequence to be bounded in order to be convergent?

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    $\begingroup$ Why does every sequence have to be either increasing or decreasing? $\endgroup$ – Randall Nov 9 '17 at 2:00
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You're proof that a sequence is convergent if and only if it is bounded is incorrect. You made the assumption that any sequence must be either increasing or decreasing. However, consider the sequence $\{a_n\} =\{1,0,1,0,\cdots\}$. This sequence is definitely bounded because $0\leq a_n\leq 1$ for all $n\in \Bbb{N}$. However, this sequence is definitely not convergent since it is oscillating infinitely between $0$ and $1$.

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