Prove if the sequence is bounded & monotonic & converges Be $a_1 = \sqrt{2} $ and $ a_{n+1} = \sqrt{2+a_n}; \forall n \in \mathbb{N}.$
1)Is this sequence {$a_n$} a monotonic function?
2)Is {$a_n$} bounded?
3)Is {$a_n$} convergent?
4)If {$a_n$} is convergent, calculate the limit of the sequence:
$\lim\limits_{n \rightarrow \infty} (a_n)$

1) $a_{n+1} = \sqrt{2+a_n}:$
For $a_2 = a_{1+1}=\sqrt{2+a_1}=\sqrt{2+\sqrt{2}}$
So for $a_n$ and $a_{n+1}:$
$a_{n+1} > a_n$ Proof: 
With $ n=1 \rightarrow  a_2>a_1$
$\sqrt{2+\sqrt{2}}>\sqrt{2}$
$ \leftrightarrow 2+\sqrt{2}>2$
So {$a_n$}  is monotonically increasing.Is all that correct?
2)Because $n \in \mathbb{N}$: $n > 0$ 
$a_1=\sqrt{2}$ is the lower bound of {$a_n$} a monotonic increasing function.
Is that correct?
3)After the Bolzano-Weierstrass theorem every monotonic bounded sequence is also convergent.
Is that enough or do I have to prove that?
4)$\lim\limits_{n \rightarrow \infty} (a_n)=...$
Now I don't really now what $a_n$ is.What shall I do?Take the limit of $a_{n+1}$?
 A: For part 1, you have only shown that $a_2 > a_1$. You have not shown that $a_{123456789} \ge a_{123456788}$, for example. And there are infinitely many other cases for which you haven't shown it either.
For part 2, you have only shown that the $a_n$ are bounded from below. You must show that the $a_n$ are bounded from above.
To show convergence, you must show that $a_{n+1} \ge a_n$ for all $n$ and that there is a $C$ such that  $a_n \le C$ for all $n$.
Once you have shown all this, then you are allowed to compute the limit.
A: The first part must be shown that the inequality holds inductively, that is, for all naturals m and n, if m>n, then a(m) > a(n). Now, you know that it is monotonically increasing.
Secondly, to show that a real sequence is bounded, you must show that there exists an M in positive reals, such that for all naturals n, |a(n)| <= M. You have merely shown the existence of a lower bound, while to show that a sequence is bounded, the existence of an upper bound is essential.
Therefore, you cannot use Bolzano-Weierstrass theorem yet, since you have not yet shown that a(n) is bounded. Moreover, your statement of this theorem is incorrect- the correct statement is that "Every bounded sequence has a convergent SUB-sequence." 
What I think you wanted to use is the property- "Every monotonically increasing sequence that is bounded above is convergent." As for the specific limit, I think you will find that if you guess the least upper bound of this sequence (probably 2, I don't know since I haven't worked anything out), that will be your limit.
