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?.

Answer

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

$$L_i=\frac{1}{1+\frac{1}{1+L_i}},\,i=1,2\Rightarrow L_i=\frac{-1+\sqrt{5}}{2}\in [0,1]$$