Converges of a sequences defined through a continued fraction

Question

Consider the following sequence $(b_n)_{n \geq 1}$ recursively given through the continued fraction

$b_1 = \frac{1}{1}$, $b_2 = \frac{1}{1+ \frac{1}{2}}$, $ \dots , b_n = \frac{1}{1+b_{n-1}}$ and show that it is convergent.

My approach: I believe to have managed to show almost all the important steps, I am stuck on the final one.

1. I managed to show that $0 < b_n \leq 1$ for all $n \in \mathbb{N} \setminus \lbrace 0 \rbrace$
2. I managed to show that $b_{2n+1} < b_{2n-1}$ and also $b_{2n+2} > b_{2n}$
3. Thus when looking at $(b_{2n})_{n \geq 1}$ I can immediatly conclude that it converges, because it is bound and monotone increasing, similarly when I look at $(b_{2n-1})_{n \geq 1}$ I can also conclude that it is convergent because it is bound and monotone decreasing.

Problem: So I define the limits $(b_{2n}) \to L$ and $(b_{2n-1}) \to M$, I want to show that $M=L$ then I am finished, because I can easily show that $(b_n) \to L=M$ follows from that: $$|L-M| = |\lim(b_{2n})- \lim(b_{2n-1})| =| \lim(b_{2n}-b_{2n-1})| $$ and I am stuck, plugging in the definition from above doesn't bring me any further.

Answer 1

Look at

$$b_{n+2} = \frac{1}{1+b_{n+1}} = \cfrac{1}{1+\cfrac{1}{1+b_n}} = \frac{1+b_n}{2+b_n}.$$

Deduce some properties of the limits $L$ and $M$.

Answer 2

$b_{2n+2} = \frac{1}{1+b_{2n+1}}, b_{2n+1} = \frac{1}{1+b_{2n}}$. Let $n\to\infty$ gives you two equations of $L,M$. Substract one from the other gives you $L=M$