Prove that the sequence $b_n = 1$, $b_{n+1} = \frac{1}{1+b_n}$ converges. I need to prove that the sequence $b_1 = 1$, $b_{n+1} = \frac{1}{1+b_n}$ converges.
If $\lim_{n\to\infty} b_n = L$ then $\lim_{n\to\infty} b_{n+1} = L$, so $\lim_{n\to\infty} b_{n+1} = \frac{1}{1+\lim_{n\to\infty} b_n} \implies L = \frac{1}{1+L} \implies L^2 + L - 1 = 0 \implies L = \frac{-1+\sqrt{5}}{2}$.
So I know that if the sequence converges to $L$ then $L = \frac{-1+\sqrt{5}}{2}$. How can I prove that the sequence converges? 
EDIT:
With the help of the answers I got my resolution:
Let's prove that $b_n$ converges by proving that the subsequences $b_{2n}$ and $b_{2n-1}$ are monotonic and bounded.
I want to prove that $\forall n \in \mathbb N$, $b_{2n} < b_{2(n+1)} < L$ and $b_{2n-1} > b_{2(n+1)-1} > L$.
I'm going to prove using induction.
First Step:
$\frac{1}{2} < \frac{3}{5} < \frac{-1 + \sqrt{5}}{2} \implies b_2 < b_4 < L $ 
and
$1> \frac{2}{3} > \frac{-1 + \sqrt{5}}{2} \implies b_1 > b_3 > L $
Second Step:
Let's prove that $b_{2n-1} > b_{2n+1} > L \implies b_{2n+1} > b_{2n+3} > L$
and $b_{2n} < b_{2n+2} < L \implies b_{2n+2} < b_{2n+4} < L$
$b_{2n-1}>b_{2n+1}>L \implies 1+b_{2n-1} > 1+b_{2n+1}>1+L \implies$
$\frac{1}{1+b_{2n-1}} = b_{2n} < \frac{1}{1+b_{2n+1}} = b_{2n+2} < \frac{1}{1+L} = L \implies $
$1+b_{2n} < 1+b_{2n+2}<1+L \implies \frac{1}{1+b_{2n}} = b_{2n +1} > \frac{1}{1+b_{2n+2}} = b_{2n+3} > \frac{1}{1+L} = L \implies$
$1+b_{2n+1} > 1+b_{2n+3} > 1+L \implies \frac{1}{1+b_{2n+1}} = b_{2n +2} < \frac{1}{1+b_{2n+3}} = b_{2n+4} < \frac{1}{1+L} = L$.
Thus $b_n$ converges.

Try showing that the subsequences of even and odd terms are monotonic and bounded. If this is the case, then both subsequences must converge. From there it shouldn't be too hard to show that they converge to the same thing, and therefore the overall sequence must converge.
