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?

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.
