Finding limit of a sequence $a_{n+1}=\frac{1}{1+a_{n}}$ For the sequence,
$a_{n+1}=\frac{1}{1+a_{n}}\; \forall n \geq 1$ , 
EDIT: $a_{1}=1$
I tried to find the monotonicity by converting it into 
$f(x)=\frac{1}{1+x}\ \implies\ f'(x)=-\frac{1}{(1+x)^2} < 0\ \forall\ x\geq 1 $ 
So the sequence is monotonically decreasing. But fiding the limit of $f(x)$ resulted in 
$\lim_{x\to\infty} f(x) = \frac{1}{1+\infty} = \frac{1}{\infty} = 0$
Is my assumption in converting the $a_{n+1}$ t0 $f(x)$ is incorrect. How can I find whether the sequence converges?. 
EDIT2: I understand that $f(x)$ formulation is incorrect. How can I prove the sequence is monotonic and find its limit?. 
 A: If you have already proved that the sequence has a limit,
you can find the limit $L$ from the recurrence equation:    
$L = \frac{1}{L+1}$     
This gives two roots but one is negative so that one cannot be the limit
(since all members of the sequence are obviously positive).     
So:    
$L = \frac{-1+\sqrt5}{2}$    
A: Note that your sequence is not monotonic, so if there is a limit, the sequence is oscillating around the limit: the even numbered terms are monotonically increasing and the odd ones are monotonically decreasing.
Sketch of the proof: 
First prove by induction that the sequence $\{a_n\}$ is positive and that $a_n\leq 1,\,\forall n\in\mathbb N$. Then show that $\{a_{2k}\},\,k=1,2,..$ is monotonically increasing and that $\{a_{2k+1}\},\,k=0,1,2,..$ is monotonically decreasing (again by induction). Because both subsequences are bounded and monotonic, they are convergent to $0\leq L_1\leq 1$ and $0\leq L_2\leq 1$, respectively. Finally you show that both limits are equal from the equation 
$$L_i=\frac{1}{1+\frac{1}{1+L_i}},\,i=1,2\Rightarrow L_i=\frac{-1+\sqrt{5}}{2}\in [0,1]$$
