Consider a sequence of real numbers $\{a_n\}$ and these 3 definitions:

(1) $\{a_n\}$ is bounded: $\exists k>0$ s.t. $-k \leq a_n\leq k$ $\forall n \in \mathbb{N}$

(2) $a_n=O(1)$: $\exists M>0, \bar{n}\in \mathbb{N}$ s.t. $-M\leq a_n\leq M$ $\forall n \geq \bar{n}$

(3) $\{a_n\}$ is convergent ($\lim_{n\rightarrow \infty} a_n=b$, $|b|<\infty$): $\forall \epsilon>0$, $\exists \bar{n}_\epsilon \in \mathbb{N}$ s.t. $-\epsilon<a_n-b<\epsilon$ $\forall n \geq \bar{n}_\epsilon$

I think that

(a) (3) implies (1). In particular, if $\{a_n\}$ is monotone then $k=b$

(b) If $\{a_n\}$ is monotone then (1) implies (3)

(c) (1) implies (2)

(d) (2) does not imply (1)

(e) (3) implies (2)

(f) (2) does not imply (3)

Are these conclusions correct? Do you see other relations among the three concepts that I have missed?


(a) "In particular" is wrong: take $a_{n}=-1+1/n$ then $b=-1$ and (potentially) every $k\ge 1$ works.

(b)-(c) Correct.

(d) Wrong: (1) and (2) are equivalent.

(e)-(f) Correct.

  • $\begingroup$ I have a question regarding (d); (2) seems to say that $\{a_n\}$ is bounded only after a certain $\bar{n}$, while (1) requires boundness $\forall n$. Hence, why you are saying (2)=(1)? $\endgroup$ – STF Nov 30 '15 at 12:28
  • $\begingroup$ I'm sorry but it is still obscure. I want to show that (2) implies (1). (2) tells me that after a certain $\bar{n}$ the sequence $\{a_n\}$ is bounded by $M>0$. We don't know what this $M$ is. How can I show that this implies that $\{a_n\}$ is bounded also for $n<\bar{n}$ by $k>0$? $\endgroup$ – STF Nov 30 '15 at 12:36
  • $\begingroup$ Because you can choose $k=\max(a_1,\ldots,a_{\overline{n}},M)$: you want to prove that $|a_n| \le k$ for all $n$. If $n\le \overline{n}$ then $a_n \le \max(a_1,\ldots,a_{\overline{n}}) \le k$. If $n>\overline{n}$ then $a_n \le M \le k$. $\endgroup$ – Paolo Leonetti Nov 30 '15 at 12:38

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