If the sequence of real numbers $(a_n)_{n=1}^{n=\infty}$ is convergent then $a_n$ is bounded My analysis textbook proves this claim as follows:
Proof: If $(a_n)_{n \in \mathbb N}$ is convergent means $\displaystyle\lim_{n=\infty} a_n =L$ which further means $$\forall \epsilon > 0,\,\exists N>0,\, \text{s.t.}\, |a_n-L|<\epsilon;\, \forall n \geq N$$.
Want to show that there exists $m,M \in \mathbb{R}$ such that $m\leq a_n \leq M$.
Since above holds for any epsilon take $\epsilon = 1$.
Then by definition there exists $N>0$ such that $|a_n-L| < 1;\, \forall n\geq N>0$.
Which means that $L-1 < a_n < L+1$. 
This is the step it differs from the way I proved it.
My version:
So we can take $m=L-1, M=L+1$ which clearly know its in the set of reals then our proof is done.
Textbook version:
Take $m=\min\{a_1,a_2,...,a_{N-1},L-1\}$ and $M=\max\{a_1,a_2,...,a_{N-1},L+1\}$
Clearly for all $n\geq N$ we have $m \leq a_n \leq M$.
My question: Why do we need to take the min and the max of $m$ and $M$? I don't really understand what those elements in the set represent either I know they are the value of the sequence at $n=1,...,N-1$ but not sure what it means.
Thanks in advance!!
 A: To show the sequence is bounded you have to find an interval that contains all the elements. The argument as far as you understand it finds an interval that contains all the elements after the $N$th, but may not contain the first $N$. There are just finitely many of those, so you enlarge it so that it does.
This is a small point just to make sure the proof matches the literal definition. All the hard work was finding an interval that covers the tail of the sequence.s
A: If the sequence converges to $L$, then a neighborhood ${\cal N}_{L}$ of $L$ (in fact, every neighborhood of $L$) contains all points of the sequence except possibly finitely many.  So the whole sequence is contained in 
$$
{\cal N}_{L} \cup (\mbox{a bounded interval that contains the remaining finitely many points}).
$$
The latter union is bounded.
A: Let $\epsilon = 1 > 0$ then there exists a positive integer $N$ such that 
$|a_n - L| < 1$ for all $n \geq N$ which implies $L-1 < a_n < L+1$ for all $n > N$
Let $M = \max({a_1, ..., a_N})$ then $ -M \leq a_k \leq M$ for $k \leq N$.
This shows that the sequence is bounded.
A: Here is something to think about, for $m = \min$, the lower bound is only true for all $n \geq N$. What about $n < N$? That's right, we have to might as well take all the finite terms in the sequence and make that lower bound. Now if perchance that any of the $|a_i| > L$, we have to take into account as well. Hence you get $M = \max\{a_1,\dots, L+1\}$.
I actually would take the absolute value of the terms to make the inequality stronger. 
A: Well, if you don't like taking the min and max of stuff, there is another way.
Let $(a_n)_{n=1}^{n=\infty}$ be any unbounded sequence.
We assume that the sequence is unbounded on the positive (right) side. It is a simple matter to constuct a subsequence $(b_n)_{n=1}^{n=\infty}$ of $(a_n)_{n=1}^{n=\infty}$ satisfying
$\tag 1 b_n \ge n \text{ for every } n \ge 1$
This subsequence does not converge. Since, in general, every subsequence of a convergent sequence must converge, $(a_n)_{n=1}^{n=\infty}$ is not a convergent sequence.
A 'mirror' argument can be used when $(a_n)_{n=1}^{n=\infty}$ is unbounded on the negative (left) side.
Any unbounded sequence either 'grows to the right' (larger and larger positive numbers) or 'grows to the left' (smaller and smaller negative numbers).
So an unbounded sequence can't be a convergent sequence.
