I need proof that the sequence defined by $a_1 =\sqrt{3}$ and $a_n = \sqrt{3 + \sqrt{a_{n-1}}} $ converges.
The proof:
Let $S = \sqrt{3 + \sqrt{3 + \sqrt{3 + ...}}}$. Then $S^2 = 3 + \sqrt{3 + \sqrt{3 + \sqrt{3 + ...}}}$. Therefore
$$S^2 - S - 3 = 0 $$
Solving this equation we easily obtain the value of S. How do I prove that the sequence is bounded? I don't understand the argument.
Induction:
Basis: $a_1=\sqrt{3}<10$
Step: Assume $a_{n-1}<10$
Proof: $a_{n-1}<10\rightarrow\sqrt{a_{n-1}}<\sqrt{10}\rightarrow 3+\sqrt{a_{n-1}}<3+\sqrt{10}\rightarrow\sqrt{3+\sqrt{a_{n-1}}}=a_n<\sqrt{3+\sqrt{10}}<10$.
Monotonic:
Basis: $a_2=\sqrt{3+\sqrt{a_1}}=\sqrt{3+\sqrt{\sqrt{3}}}>\sqrt{3}=a_1$
Step: Assume $a_{n-1}>a_{n-2}$
Proof: $a_{n-1}>a_{n-2}\rightarrow \sqrt{a_{n-1}}>\sqrt{a_{n-2}}\rightarrow 3+\sqrt{a_{n-1}}>3+\sqrt{a_{n-2}}\rightarrow
\\
\rightarrow a_n=\sqrt{3+\sqrt{a_{n-1}}}>\sqrt{3+\sqrt{a_{n-2}}}=a_{n-1}$
So we got that the sequence is monotonic and bounded, so is convergent. From here:
$\lim_{n\to\infty} a_n=\lim_{n\to\infty}\sqrt{3+\sqrt{a_{n-1}}}\rightarrow \lim_{n\to\infty} a_n=\sqrt{3+\sqrt{\lim_{n\to\infty}{a_{n-1}}}}\rightarrow
S=\sqrt{3+\sqrt{S}}\rightarrow \\
\rightarrow S^2=3+\sqrt{S}\rightarrow S^4-6S^2+9=S\rightarrow S=2.11\backslash 1.35$
We know that $1.35<a_1<a_2<\cdots<a_n$.
So it cant be the answer. Which means our limit is(about) $2.11. \square$
