Computing the first few terms $$a_1=1, a_2=\sqrt{3}=1.732....,a_3=1.9318....,a_4=1.9828...$$ I feel that $(a_n)_{n\in \mathbb{N}}$ is bounded above by 2, although I have no logical reasoning for this. Since, $(a_n)_{n\in \mathbb{N}}$ is monotone increasing sequence, it must converge by monotone convergence theorem, and converge to 2.
Can anyone help me to make this more formal? Besides, I would really appreciate if anyone could shed some light on how to find the bound and limit of such sequences (that are not in closed form but in recursion).
Answer
Hints:
$$a_1\le a_2\;\;\text{and}\;\;a_{n+1}:=\sqrt{2+a_n}\stackrel{\text{induction}}\le\sqrt{2+a_{n+1}}=:a_{n+2}$$
$$a_1\le 2\;\;\text{and}\;\; a_{n+1}:=\sqrt{2+a_n}\stackrel{\text{induction}}\le\sqrt{2+2}=2$$
The above shows your sequence is a monotone ascending one and bounded above, so its limit exists, say it is $\;w\;$, and now a little arithmetic of limits:
$$w\leftarrow a_{n+1}=\sqrt{2+a_n}\rightarrow\sqrt{2+w}$$
so $\;w=\ldots\ldots?$
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