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 $e_n$?
 A: 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 :

A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that
  $\ds{e_{n + 1} = {e_{n} \over e_{n} + 2} \implies
{1 \over e_{n + 1}} + 1 = 2\pars{{1 \over e_{n}} + 1} \implies
\bbx{\ds{{1 \over e_{n + 1}} + 1 = 2^{n + 1}\pars{{1 \over e_{0}} + 1}}}}$

With the above expression we can discuss several $\ds{e_{n}}$-behaviours !!!.
A: show with induction that for your sequence is hold $$e_n=\frac{2}{c_1 2^n+2^{n+1}-2}$$
A: When you want to study a homographic sequence $u_{n+1}=\frac{au_{n}+b}{cu_{n}+d}$, start by solving the equation $x=\frac{ax+b}{cx+d}$:


*

*if you get two roots $y$ and $z$, the sequence $v_{n}=\frac{u_{n}-y}{u_{n}-z}$ is geometric;

*if there's only one root w, the sequence $v_{n}=\frac{1}{u_{n}-w}$ is arithmetic.

