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.

Here's another interesting way of proving its convergence (And might provide some nice insights into the limit): Define
$$b_n=\frac{h_n}{g_n}$$
where these are the numerator and denominator of $b_n$ (with no reductions to a simpler form after appyling the recursion formula). Then from your definition
$$b_{n+1}=\frac{h_{n+1}}{g_{n+1}}=\frac{1}{1+\frac{h_n}{g_n}}=\frac{g_n}{h_n+g_n}.$$
Hence,
$$h_{n+1}=g_n$$
and
$$g_{n+1}=h_n+g_n.$$
Rearranging slightly, we get
$$h_{n+1}=h_n+h_{n-1}$$
and
$$g_{n+1}=g_n+g_{n-1}.$$
Of course, we can find just by computing the first few terms that $h_1=h_2=1$, $g_1=1$, and $g_2=2$. However, this is simply the recursion formula for the fibonacci numbers! Thus, we know $h_n=F_n$ and $g_n=F_{n+1}$ (where $F_n$ is the $n$th fibonacci number). Therefore,
$$b_n=\frac{F_n}{F_{n+1}}.$$
At this point, there are many ways to show that the limit exists (And also show why you got the limit of $\frac{-1+\sqrt{5}}{2}$.).
