Proof that bounded growth of a sequence implies convergence I have to prove that:

If a positive sequence of numbers $a_n$ is growing and bounded above, then the sequence $a_n$ converges to some finite value as $n$ increases.

Here is my 'proof' so far:
Suppose that a positive sequence $a_n$ is growing such that $a_n < a_{n+1}$, and bounded above by some number A such that $a_n \leq A, \forall n \in \mathbb{N}$.
We know from the supremum axiom that the set $M = {a_1, a_2, a_3, ...}$ not only has the upper bound A but a smallest upper bound which we can denote $\sup M = L$
Let $\epsilon > 0$ be a fixed and unknown number. We want to prove that $a_n \rightarrow L$, and we do this by finding a number $N$ so that 
$$n>N \implies |a_n - L| < \epsilon$$
Opening up the absolute value sign we get: $$L - \epsilon < a_n < L + \epsilon$$
Since $L = \sup M$ of the set $M$ which contains all elements of the sequence $a_n$ and since $\epsilon > 0$ it follows that there must exist some $N$ such that $L-\epsilon < a_N$, which means that for all $n>N$ we have $L - \epsilon < a_n$.

Two questions: does my proof look alright so far? And secondly, I am supposed to find a suitable choice for $N$ which makes the implication $n>N \Rightarrow |a_n - L| < \epsilon$ true. How do I do this when I don't know what the sequence looks like?
 A: There is an error of logic in your proof.
The statement $$n>N \implies |a_n - L| < \varepsilon$$ is exactly what you want to prove to show that $a_n$ converges to $L$. What you have done is use this statement to prove the very same statement.
There is no way you can find a fixed relationship between $N$ and $\varepsilon$ by the way, as you have a generic and unspecified monotone increasing sequence. You can however do the proof nonetheless.
Hint: There are many different ways of proving this, but one fundamental approach could be: for a particular $\varepsilon$ and some integer $m>1$, consider the set $[L-\varepsilon, L-\varepsilon/m)$ and $[L-\varepsilon/m, L]$. Where do an infinite number of elements $a_i$ reside? Then increase $m$.
A: The proof looks basically ok. The idea is certainly correct.
I would write, instead of "let $\epsilon$ be a fixed and unknown number", something like "choose an arbitrary $\epsilon>0$". I would also perhaps point out that by definition of supremum we automatically and trivially have $a_n <L+\epsilon$ for all $n$. And I'd similarly point out that the last conclusion (if it works for $a_N$, it works for every $a_n$ with $n>N$) follows from the fact that the sequence is non-decreasing. The last two observations are probably superfluous if writing "professionally", but it's probably better to write them down for a homework.
To answer your second question: how can I know which N is "sufficiently large" for a given $\epsilon$? Well, as we said, if $a_N > L-\epsilon$, you are ok for every $n>N$ too, since the sequence is non-decreasing.  Special sequences might sport special properties that allow you to determine a "sufficiently large N" but, in general, determining $N$ might require you to try $a_1$, $a_2$ etc. until you find something exceeding $L-\epsilon$.
