Find limit of $a_{n+1} = \frac{2a_n}{a_n+1}$. 
Determine if the following sequence $(a_n)_{n \in \mathbb{N}}$ converges and if so, find the limit. $a_1 \geq 0$, $a_{n+1} = \frac{2a_n}{a_n+1}$.

What I've done so far:
First we can notice that if $a_1 = 0 \Rightarrow a_n = 0, n \in \mathbb{N}$ and thus $\lim\limits_{n \to \infty} a_n=0$.
If $a_1 > 0, a_n > 0$ by induction and $a_{n+1} = 2 - \frac{2}{a_n + 1}$ which gives $a_{n+1}a_n + a_{n+1}-2a_n = 0$. I'm not familiar with recurrence equations, but WolframAlpha gives an equation such that a limit of it is easy to evaluate and equals $1, n \rightarrow \infty$.
Is there another way to find the limit of this sequence?
 A: Let $f(x) = \frac{2x}{x+1}$ for $x\geq 0$, so that $a_{n+1}= f(a_n)$.
Then, we have :
$$\forall x\geq 0, f(x)-x = \frac{x(1-x)}{x+1}$$
Therefore $1>f(x) > x$ when $x\in (0,1)$ and $1<f(x)<x$ when $x>1$.

*

*If $a_1$ is equal to $0$ or $1$, then $f(a_0) = a_0$ and the sequence is constant.


*If $a_1 \in (0,1)$, then you can show by induction that for all $n\in \mathbb N$,
$$0<a_n<a_{n+1}<1$$
Therefore, $(a_n)$ converges to some $\ell \in (0,1]$. Taking the limit in $a_{n+1} = f(a_n)$, we find $\ell = f(\ell)$, whose only solution on $(0,1]$ is $\ell = 1$.


*If $a_1 \in (1,+\infty)$, then by induction :
$$\forall n\in\mathbb N, 1<a_{n+1}<a_n $$
Therefore $a_n \to \ell \geq 1$. Again we have $f(\ell) = \ell$ which implies $\ell =1$.
Conclusion

If $a_1 = 0$, then $(a_n)$ converges to $0$. If $a_1>0$, then $(a_n)$ converges to $1$.

A: If $a_1\neq 0$, then $a_n\neq 0$ for all $n$ (prove this), so we have $\dfrac{1}{a_{n+1}} = \dfrac{1}{2}\dfrac{1}{a_n} + \dfrac{1}{2}$, or $\dfrac{1}{a_{n+1}} - 1 = \dfrac{1}{2}\Big(\dfrac{1}{a_n} -1\Big)$. Hence $\dfrac{1}{a_n} - 1 = \dfrac{1}{2^{n-1}}\Big(\dfrac{1}{a_1} -1\Big)$. Now what can you say about $\displaystyle{\lim_{n\to\infty}} a_n$?
Remark: the recurrence $x_{n+1} = \dfrac{a x_n+b}{c x_n+d}$ (where $a,b,c,d\in\mathbb{C}$ satisfy $ad-bc\neq 0$, this is called a Mobius transformation) always implies that $\{x_n\}$ has an elementary general formula.
A: There's a related proof to above using Fixed Points. Let $f(x)=2x/(x+1)=2-2/(x+1)$.
A Fixed Point is by definition a solution of $f(x)=x$.
Solving $x=2x/(x+1)=0=x^2-x \implies x=1$ or $x=0$.
$f'(x)=\frac{2}{(1+x)^2}$. In particular $f'(1)=1/2$.
Let $0<x<y$.
$f(x)-f(y)=2(\frac{1}{y+1}-\frac{1}{x+1})=2\frac{(x-y)}{(x+1)(y+1)}\implies |f(x)-f(y)|<|x-y|$
So if $f(y)=y$, then $|f(x)-y|<|x-y|$, so $x_n\to y$ for each iteration.
Further, for differentiable functions, $|f'(y)|<1$ also implies iterations get closer together and converge to y.
This provides a way to find limits: convert $a_{n+1}=f(a_n)$ to $f(x)=x$. Solve for $x$ and choose from those solutions where $f'(x)\le 1$.
