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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 $a_{n+1}:$
$a_{n+1} > a_n$

With $ n=1 \rightarrow a_2>a_1$
$ \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 $a_{n+1}$?

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1 Answer 1

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.

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