Show that an increasing sequence diverges if and only if it is unbounded. Show that an increasing sequence diverges if and only if it is unbounded.
How should I go about proving this?
 A: If it's unbounded, the sequence diverges. This is clear.
So you only need to show that a bounded, increasing sequence converges. This is a well-known result often called the Monotone Convergence Theorem. In short, call the supremum of the sequence $L$. Then $L$ is the limit. In particular, for any $\epsilon > 0$, there is some $N$ such that $a_n$ satisfies $L - a_n < \epsilon$, for all $n > N$.
A: Any unbounded increasing sequence necessarily goes off to $\infty$, so perhaps the only nontrivial direction here is showing that every increasing, divergent sequence is unbounded. The contrapositive is to show that every bounded increasing sequence is convergent. But this is exactly the statement of the Monotone Convergence Theorem (http://en.wikipedia.org/wiki/Monotone_convergence_theorem).
A: In the "if" direction: Try to see how undboundedness interferes with the definition of a limit.
In the "only if" direction: Suppose the increasing sequence is in fact bounded (say, with least upper bound $M$.) Is there any way it could possibly not converge?
A: Let $\{a_n\}_{ n \in \Bbb N}$ be increasing function and bounded , if $a = \sup a_n$ then $\lim_{n \to \infty} a_n = a$ because for every $\epsilon >0$ there exist $N \in \Bbb N$ such that for every $n \geq N$ $$a-\epsilon < a_n < a< a+\epsilon$$ and this is cntradiction
Noe let $\{a_n\}_{ n \in \Bbb N}$ be a unbunded and converge to $a$ then for every $\epsilon >0$ there exist $ N \in \Bbb N$ sucht that for every $n \geq N$ $$ |a_n|<|a|+ \epsilon$$
If $b= \max\{a_1,\cdots ,a_{N-1}\}$ we clami that $$\forall n \in \Bbb N, |a_n|< \max \{b,|a|+ \epsilon\} $$
Which is contradiction.
