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?

  • 4
    $\begingroup$ Why does every sequence have to be either increasing or decreasing? $\endgroup$ – Randall Nov 9 '17 at 2:00

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 $1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.