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 $e_n$?

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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 !!!.
