Convergence of $\sqrt{3 + \sqrt{3 + \sqrt{3 + ...}}}$

Question

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.

Your "$a_n = \sqrt{3 + \sqrt{a_{n-1}}}$" contradicts $S = \sqrt{3 + \sqrt{3 + \sqrt{3 + ...}}}.$ Don't you rather mean $a_n = \sqrt{3 +a_{n-1}}$? If (and only if) so, the links in the closure banner would answer your question. — Anne Bauval, Nov 06 '23 at 18:06

score 1 · Accepted Answer · answered Sep 23 '23 at 21:06

1

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) $2.11. \square$

answered Sep 23 '23 at 21:06

Cusp Connoisseur

600

That! I was trying to prove what you did. Thank you very much. – Gleberson Antunes Sep 24 '23 at 04:12
I suspect a misprint in the question. Once the question corrected, this answer would not be appropriate any longer, and the links in the closure banner would suffice. – Anne Bauval Nov 06 '23 at 18:09
what do you mean by misprint? what do you think @GlebersonAntunes tried to write? – Cusp Connoisseur Nov 06 '23 at 18:10
Ok @AnneBauval I see what you mean. that $a_1=3$ and not $\sqrt{3}$? I don'tthink this is a big deal because he wanted to know how to prove convergence and he didnt ask to find the limit – Cusp Connoisseur Nov 06 '23 at 18:12
No, I did not mean that. I meant the recurrence relation does not correspond to what we are used to. See my comment above. – Anne Bauval Nov 06 '23 at 18:15
I see what you mean, but the misprint is in his answer, not in the question. He tried(with S) to calculate the sum, but his main question was how to proof its convergence - the question is already(kinda) closed, because it's a duplicate. – Cusp Connoisseur Nov 06 '23 at 18:21
It was closed by error @Chessplayer: it would be a duplicate (and A.Goodier's comment above would be correct) if the question was modified as I suggest (and as I suspect the OP received it but misread it), with $a_n = \sqrt{3 +a_{n-1}}$ instead of $a_n = \sqrt{3 +\sqrt{a_{n-1}}}.$ – Anne Bauval Nov 06 '23 at 20:11
Let us continue this discussion in chat. – Cusp Connoisseur Nov 06 '23 at 20:26

Convergence of $\sqrt{3 + \sqrt{3 + \sqrt{3 + ...}}}$

1 Answers1