Prove $a_n=1+\frac{a_{n-1}}{1+a_{n-1}}$ increasing There is a homework question in Calculus-1 course:
Calculate the limit of $\{a_n\}$: $$a_1=1,\ a_n=1+\frac{a_{n-1}}{1+a_{n-1}}$$
I think the key points are bounded and increasing, and I have proved that $$a_n\in(1, 2)$$ 
If I knew it's increasing then $$a=1+\frac{a}{1+a}\Rightarrow\lim a_n=\frac{\sqrt5+1}{2}$$
My question is How to Prove it's increasing?
I tried it in two ways:
$$a_{n+1}-a_n=1+\frac{a_n}{1+a_n}-a_n=\frac{-a^2_n+a_n+1}{1+a_n}$$
But how to prove that $-a^2_n+a_n+1>0$?
Another way is
$$\frac{a_{n+1}}{a_n}=\frac{1}{a_n}+\frac{1}{1+a_n}=\frac{1+2a_n}{a_n+a^2_n}$$
But how to prove that $1+2a_n>a_n+a^2_n$?
This is not a proof question which means $\frac{\sqrt5+1}{2}$ is not a known result.
Thank you!

First, we prove that $a_n$ is bounded above. Obviously, $a_n = 1+ \frac{a_{n-1}}{a_{n-1}+1} < 2$. 
To prove that the sequence is increasing, we can just use induction. The base cases are trivial so suppose the claim holds for all naturals $ \le k$. Then $a_{k+1}-a_k = \frac{a_k}{1+a_k} - \frac{a_{k-1}}{1+a_{k-1}} = \frac{a_k(1+a_{k-1})-a_{k-1}(1+a_k)}{(1+a_k)(1+a_{k-1})}$.
The numerator is $a_k-a_{k-1}$ which by the inductive  hypothesis is $> 0$ so we are done.
