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?
_
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: Solution $\blacktriangleleft$. Let 
$$
f(x) = \frac x 2 + \frac 1 {1+x}, \quad [x >-1],
$$
then 
$$
f'(x) = \frac 12 - \frac 1{(1+x)^2} = \frac {(x+1+\sqrt 2)(x+1-\sqrt 2)} {2(1+x)^2}, 
$$
hence $f(x) \searrow$ on $(-1, \sqrt 2 -1)$, $\nearrow$ on $(\sqrt 2 - 1, +\infty)$, therefore $f(x) \geqslant f(\sqrt 2 -1) = \sqrt 2 -1/2$.  Thus for $n \in \Bbb N^*$, $a_n \geqslant \sqrt 2-1/2.$
Now for $n \geqslant 1$, 
\begin{align*}
a_{n+1}-a_n &= \frac {a_n -a_{n-1}}2 + \frac 1{1+a_n} - \frac 1{1+a_{n-1}} \\
&= (a_n -a_{n-1}) \left(\frac 12 - \frac  1 {(1+a_n) (1+a_{n-1})}\right). 
\end{align*}
Since 
$$
\frac 12 > \frac 12-\frac 1 {(1+a_n)(1+a_{n-1})} \geqslant \frac 12 - \frac 1 {(\sqrt 2 + 1/2)^2} = \frac 12 - \frac 1 {2 + 1/4 +\sqrt 2} >0, 
$$
we have 
$$
|a_{n+1} - a_n| = |a_{n-1} - a_n| \left|\frac 12 - \frac 1{(1+a_n)(1+a_{n-1})}\right| \leqslant \frac 12 |a_n - a_{n-1}|,
$$
hence $(a_n - a_{n-1})$ is contracting, thus $(a_n)$ is Cauchy. Hence $(a_n)$ converges. Since $a_n \geqslant \sqrt 2 -1/2 >0$, we have $a_n \to 1. \blacktriangleright$
A: $$a_0>-1$$
$$a_{n+1}=\dfrac{a_n}{2}+\dfrac{1}{1+a_n}=\dfrac{a_n^2+a_n+2}{2(1+a_n)}$$
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: $a_{n+1}=\dfrac{a_n^2+a_n+2}{2(1+a_n)}<1, \blacksquare .$
$$a_{n+1}-a_n=\dfrac{a_n^2+a_n+2}{2(1+a_n)}-a_n=\dfrac{a_n^2+a_n+2-2a_n-2a_n^2}{2(1+a_n)}=-\dfrac{a_n^2+a_n-2}{2(1+a_n)}=$$
$$=-\dfrac{(a_n+2)(a_n-1)}{2(a_n+1)}>0,\forall n\geq 1\rightarrow \{ a_n\} \, \text{is increasing and bounded.}$$
$$L=\dfrac{L}{2}+\dfrac{1}{1+L}\rightarrow L=\dfrac{L^2+L+2}{2(1+L)}\rightarrow 2L^2+2L=L^2+L+2\rightarrow$$
$$L^2+L-2=0\rightarrow L=-2\quad \text{or}\quad \boxed{L=1}$$
