How to show sequence defined by this recursive formula converges to 0? As part of the problem I'm working on, I reached the point where I have to show the sequence of error terms $e_n$ defined by:
$$ e_{n+1} = \frac{e_n}{e_n+2} $$
converges to 0 for choice of initial $e_0 > -1$
I've been able to show this for $e_0 \geq 0$, as $e_n \geq 0 \implies 0 < e_{n+1} \leq \frac{1}{2}e_n$
How can one show that convergence to $0$ still holds for $-1 < e_0 < 0$?
Is there a way to prove this using only the non-explicit definition of $e_n$?

Suppose $x\in(-1,0)$. We have $x+2>1$, thus $\frac1{x+2}<1$ hence $\frac x{x+2}>x$ and still $\frac x{x+2}\in(-1,0)$.
Combined with induction, this leads to :
If $e_0\in(-1,0)$, then $\forall n\in\mathbb{N},e_n\in(-1,0)$ and $e_{n+1}>e_n$.
The sequence $(e_n)$ is well defined, strictly increasing and has an upper bound, hence converges to some fixed point of the continuous map $f:[-1,0]\to[-1,0],x\mapsto\frac x{x+2}$. There are only two fixed points : $-1$ and $0$, but the sequence $(e_n)$ cannot converge to $-1$ (since $e_0>-1$ and it is an increasing sequence); so it converges to $0$.
Here is an illustration :

