limit of the sequence $a_n = \sqrt{2+\sqrt{3+\sqrt{2+\sqrt{3+...}}}}$

Question

I have already shown that the sequence is monotone and bounded, so it is convergent. what I need is to calculate the limit when $a_n = \sqrt{2+\sqrt{3+\sqrt{2+\sqrt{3+...}}}}$, $n \rightarrow \infty$, unnder the condition "do not use polynomials of degree 4" to calculate said limit. I don't know the reason why the teacher put that condition

Answer 1

This is the only way I can think of to get a solution that doesn't involve polynomials of degree $4$.

The following system of equations gives the solution of the problem: $\left\{\begin{matrix}a^2=2+b\\b^2=3+a\end{matrix}\right. $, in particular, the solution to $a$ would give the desired result. This is because $a=\sqrt{2+\sqrt{3+\sqrt{2+\cdots}}}$ and $b=\sqrt{3+\sqrt{2+\sqrt{3+\cdots}}}$ Therefore, $a=\sqrt{2+b}$ and $b=\sqrt{3+a}$. And it is possible to solve that system numerically, by Newton's method, for example (numerical methods for this type of problems are usual, because many times the analytical solutions are horrendous).

Answer 2

This does not answer the question but is here to show the connection to quartic polynomials for the OP. I'm guessing there is some work around to not have to solve the quartic.

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Let $a=\lim_{n\to\infty}a_n$ denote the number in question, namely, $$ a = \sqrt{2+\sqrt{3+\sqrt{2+\sqrt{3+...}}}} $$ The "trick" in these types of problems (as pointed out in the comments) is to notice that the right hand side can also be written in terms of $a$ yielding $$ a=\sqrt{2+\sqrt{3+a}}. $$ With some simple algebraic manipulations this result can be recast as the quartic equation $$ a^4-4a^2-a+1=0. $$ This quartic has four real roots. To determine which root we want we note that $a>\sqrt 2=1.41\dots$, which leaves us with only root that fits these criteria $$ a=\frac{1}{2} \sqrt{\frac{1}{3} \left(8+\frac{28}{\sqrt[3]{\frac{1}{2} \left(187+3 i \sqrt{5871}\right)}}+\sqrt[3]{\frac{1}{2} \left(187+3 i \sqrt{5871}\right)}\right)}+\frac{1}{2} \sqrt{\frac{16}{3}-\frac{28}{3 \sqrt[3]{\frac{1}{2} \left(187+3 i \sqrt{5871}\right)}}-\frac{1}{3} \sqrt[3]{\frac{1}{2} \left(187+3 i \sqrt{5871}\right)}+\frac{2}{\sqrt{\frac{1}{3} \left(8+\frac{28}{\sqrt[3]{\frac{1}{2} \left(187+3 i \sqrt{5871}\right)}}+\sqrt[3]{\frac{1}{2} \left(187+3 i \sqrt{5871}\right)}\right)}}}=2.061\dots $$