This time we consider some abstract results as what sequences converges.

• monotone bounded sequence converges.
• what is $\limsup$ and $\liminf$? The $\epsilon$-of-room trick.
• Cauchy sequence.

Discussion time: Ex 10.1, 10.6 in Ross

This is in Definition 9.8. Informally, it says for any $M$ (no matter how large it is), the sequence will eventually stay above $M$.

A sequence that is (weakly) increasing, namely $a_{n+1} \geq a_n$. Examples: (a) $a_n = n$. (b) $a_n = 1 - n$. © $a_n = \log n$.

Theorem: bounded monotone sequence is convergent. How do we construct such a limit? The limit is obviously the 'largest' number, but there is no largest number in the sequence. We should use 'sup' to get the largest number. But sup needs to applied to a set, not a sequence, so we need to first convert a sequence to its 'value set', $S = \{a_n \mid n \in \N\}$, and define $a = \sup S$. How do we show that $a_n \to a$? We need to show that for any $\epsilon > 0$, we have $|a_n - a| < \epsilon$ for all large enough $n$. Let's first show that, there exists one $n_0$, such that $|a_{n_0} - a| < \epsilon$, this holds by definition of the $\sup S$, namely, there is one element in $S$ that satisfies this condition, it must be $a_{n_0}$ for some $n_0$. Then, we can use monotone condition to say, for all $n > n_0$, we have $a_n \geq a_{n_0}$. Furthermore, since $a_n \leq a$, we know $a_n \in [a_{n_0} , a) \In (a-\epsilon, a)$ for all $n > n_0$, then we are done.