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.
Best Answer
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 $a_n \to 1. \blacktriangleright$