Convergence and limit of recursive sequence Consider the sequence $(a_n)_{n \in \mathbb{N}}$ which has startvalue $a_0 > -1$ and recursive relation:
$$a_{n+1} = \frac{a_n}{2} + \frac{1}{1+ a_n}$$
How to prove the convergence and find the limit?
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I think you need to show the convergence with a Cauchy sequence. It is also possible to show that the sequence is decreasing and bounded, but I found that if $-1 < a_0 < 0$ that the sequence first increases and after $a_n > 1$ it starts to decrease. So that method seems more complicated to me. 
For the limit I found the values $1$ and $-2$ but don't know how to show which is the right one. 

$$a_0>-1$$
$$a_{n+1}=\dfrac{a_n}{2}+\dfrac{1}{1+a_n}=\dfrac{a_n^2+a_n+2}{2(1+a_n)}$$
We know that $x^2+x+2>0, \forall x\in \mathbb{R}$.
$$a_1=\dfrac{a_0^2+a_0+2}{2(1+a_0)}>0\rightarrow a_{n+1}=\dfrac{a_n^2+a_n+2}{2(1+a_n)}>0\rightarrow \boxed{a_n>0, \forall n\geq 1}$$
We are going to prove by induction that $\boxed{a_n<1,\forall n\geq 1}$
If $x<1\rightarrow \dfrac{x^2+x+2}{2(1+x)}<1$
Then: $a_{n+1}=\dfrac{a_n^2+a_n+2}{2(1+a_n)}<1, \blacksquare .$
$$a_{n+1}-a_n=\dfrac{a_n^2+a_n+2}{2(1+a_n)}-a_n=\dfrac{a_n^2+a_n+2-2a_n-2a_n^2}{2(1+a_n)}=-\dfrac{a_n^2+a_n-2}{2(1+a_n)}=$$
$$=-\dfrac{(a_n+2)(a_n-1)}{2(a_n+1)}>0,\forall n\geq 1\rightarrow \{ a_n\} \, \text{is increasing and bounded.}$$
$$L=\dfrac{L}{2}+\dfrac{1}{1+L}\rightarrow L=\dfrac{L^2+L+2}{2(1+L)}\rightarrow 2L^2+2L=L^2+L+2\rightarrow$$
$$L^2+L-2=0\rightarrow L=-2\quad \text{or}\quad \boxed{L=1}$$
