I was studying the topic of recursively defined sequences and stumbled upon the following question. Let's say we have a sequence $(a_n)_n$ which meets the following conditions: $a_0 > -1$ and $$a_{n+1} = \frac{a_n}{2} + \frac{1}{1+a_n}$$ for every $n \in \mathbb{N}$. We need to proof that the sequence converges. Now I approached this problem by turning the equation into the following function $$F:\mathbb{R} \rightarrow \mathbb{R}: x \rightarrow \frac{x}{2} + \frac{1}{1+x}$$ I then differentiated the function and searched the supremum of $|F'(x)|$ for x > -1 and hoped this would give a value smaller then 1 so that the function would be a contraction and thus due to the theorem of Banach the sequence would converge, but to my dissapointment this did not work. I know the function is a contraction for some values of x but not all values of x > -1.
Does anyone have tips for approaching this problem and maybe approaching it with a different methode. Any help would be greatly appreciated :))
Alright so if
$$f:(-1,\infty)\to\mathbb{R},\quad x\mapsto\frac{x}{2}+\frac{1}{1+x},$$
then $f'$ is given by
$$f'(x)=\frac{1}{2}-\frac{1}{(1+x)^2}.$$
Now if $x\geq0$, then
$$\lvert f'(x)\rvert=\left\lvert\frac{1}{2}-\frac{1}{(1+x)^2}\right\rvert\leq\frac{1}{2}.$$
Thus, using the MVT and taking the supremum as you yourself suggested, we find that $f$ is a contraction on $[0,\infty)$. Now it is easy to check that if $x\in(-1,\infty)$, then $f(x)\in[0,\infty)$. In particular this means that $a_j\in[0,\infty)$ for all $j\geq1$. This means that $f$ is indeed a contraction where every entry of the sequence (with the possible exception of $a_0$) is, and so we can apply the Banach fixed point theorem to get that $\{a_j\}_{j\in\mathbb{N}}$ converges.
If you also wish to find the limit, just solve $f(x)=x$, and $x$ is then your limit.