Finding the limit of the sequence $(1, \frac{1}{2+1}, \frac{1}{2+\frac{1}{1+1}},\cdots)$

Question

I am struggling to find the limit of the sequence:

$$1, \cfrac{1}{2+1}, \cfrac{1}{2+\cfrac{1}{1+1}}, \cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+1}}}, \cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+1}}}},\cdots$$

Normally, for these types of sequences, I would find a recurrence relation, i.e. $a_n=f(a_{n-1})$, and then say as $n\to\infty$, $a=f(a)$ which I would solve for $a$. I cannot find a recurrence relationship between each term in this sequence, however. What is the recurrence relationship between the terms in this sequence? Alternatively, is there another way of finding its limit?

Answer 1

Based on the first five terms I got the following:

$a_1=1$, $\hspace {0,2cm}$ $a_2=\frac{1}{3}$, $\hspace {0,2cm}$

$a_3=\frac{1}{2+\frac{1}{1+a_1}}$

$a_4=\frac{1}{2+\frac{1}{1+a_2}}$

$a_5=\frac{1}{2+\frac{1}{1+a_3}}$

So the recurrence relationship $\hspace {0,2cm}$ $a_n=\frac{1}{2+\frac{1}{1+a_{n-2}}}=\frac{1+a_{n-2}}{3+2a_{n-2}}$

The limit of the sequence can be easily computed by the following way:

If the limit existing and $\lim\limits_{n\rightarrow\infty}= L>0$

then solving the $\hspace {0,2cm}$ $L=\frac{1+L}{3+2L}$ eqution we get that $L=\frac{\sqrt{3}-1}{2}$

(Sorry for the delayed ansver)