while studying the topic of recursively defined sequences I stumbled upon the following problem which I got stuck with, it goes as follows. Let $(b_n)_n$ be a sequence for which the following recursion formula holds $$b_{n+1} = \frac{b_n}{2} - \frac{1}{1+b_n}$$ for every $n \in \mathbb{N}$. I now need to find every starting value for which this sequenced converges.
Now the first thing i did was to see if there were any fixed points, but to my dissapointment i couldn't find any. I don't know any other methode on how to start solving this problem, so if anybody could maybe give a tip on how to start with this problem or give a method for solving it it would be greatly appreciated :))
Generally speaking, suppose you're studying a recursion of the form $b_{n+1} = f(b_n)$ where $f : \mathbb{R} \to \mathbb{R}$ is continuous. Then taking limits on both sides gives, by continuity,
$$\lim_{n \to \infty} b_{n+1} = L = \lim_{n \to \infty} f(b_n) = f(L)$$
so the limit, if it exists, must be a fixed point of $f$. So if $f$ does not have a fixed point, this sequence cannot have a (finite) limit. A simple example here is $f(x) = x + 1$.
In this case we have $f(x) = \frac{x}{2} - \frac{1}{1+x}$. The fixed point equation is $f(x) = x \Leftrightarrow x^2 + x + 2 = 0$, which has no real roots. So there are no fixed points, as you've said, and hence no starting value for which the sequence converges. (Strictly speaking $f(x)$ has a pole at $x = -1$ but this makes the analysis simpler rather than more complicated: if the sequence $b_n$ ever takes the value $-1$ then it's never defined afterwards.)
If there had been fixed points (for example, there are fixed points when $f(x) = \frac{x}{2} + \frac{1}{1+x}$), the next thing you'd need to do would be to compute the basin of attraction of each fixed point. In nice cases you can do this by playing around with derivatives.