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

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)
