Convergence of sequence to $\to \dfrac{a+b\sqrt c}{d} ,a,b,c,d \in\mathbb{N}$

Question

$$\sqrt[8]{2207-\dfrac{1}{2207-\dfrac{1}{2207-\dfrac{1}{2207\ddots}}}} $$converges to $$\to \dfrac{a+b\sqrt c}{d}$$and,we know $$a,b,c,d \in \mathbb{N}$$ find $a,b,c,d$
I tried to find a recursive relation for ${2207-\dfrac{1}{2207-\dfrac{1}{2207-\dfrac{1}{2207\ddots}}}}$
I got $$2207 -(1/x)=x$$ then find $x$ but finding $x $ in form of $\dfrac{a+b\sqrt c}{d}$ is like a challenge.

but get stuck in calculations ,my sense is to do a tricky way for this problem !

I am thankful of your hint ,in advanced

Answer 1

Let $c$ be your continued fraction. Then $$c=2207-\frac1c$$ that is $$c^2-2207c+1=0$$ giving $$c=\frac{2207+987\sqrt{5}}{2}.$$ We find $$c^{1/2}=\frac{47+21\sqrt{5}}{2},$$ $$c^{1/4}=\frac{7+3\sqrt{5}}{2}$$ and $$c^{1/8}=\frac{3+\sqrt5}2.$$

Answer 2

let the big thing be x. Then $2207 -(1/x)=x.$ Then solve x using quadratic formula. The final answer is x^(1/8)

Answer 3

Let $a_1=2207$ and $\displaystyle a_{n+1}=2207-\frac{1}{a_n}$. Then

$$a_{n+1}-a_n=\frac{1}{a_{n-1}}-\frac{1}{a_n}=\frac{a_n-a_{n-1}}{a_na_{n-1}}$$

So $a_2<a_1$ would imply that $a_{n+1}<a_n$ for all $n\in\mathbb{N}$.

$\{a_n\}$ is a strictly decreasing sequence which is bounded from below. So it is convergent. Let $a$ be the limit. Then

$$a=2207-\frac{1}{a}$$

Let $b=a^{1/8}$. Then $a=b^8$ and

\begin{align} \left(b+\frac{1}{b}\right)^2&=b^2+\frac{1}{b^2}+2\\ \left(b^2+\frac{1}{b^2}\right)^2&=b^4+\frac{1}{b^4}+2\\ \left(b^4+\frac{1}{b^4}\right)^2&=b^8+\frac{1}{b^8}+2\\ \left(b^4+\frac{1}{b^4}\right)^2&=2207+2=2209\\ b^4+\frac{1}{b^4}&=47\\ \left(b^2+\frac{1}{b^2}\right)^2&=49\\ b^2+\frac{1}{b^2}&=7\\ \left(b+\frac{1}{b}\right)^2&=9\\ b+\frac{1}{b}&=3\\ b^2-3b+1&=0 \end{align}

$a_1>1$ and $\displaystyle a_{n+1}=2207-\frac{1}{a_n}$ would imply that $a_n>1$ for all $n\in\mathbb{N}$ (by induction). So $a\ge 1$ and hence $b\ge 1$. Therefore,

$$b=\frac{3+\sqrt{5}}{2}.$$