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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 $e_n$?

Manuel Guillen
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  • I think you can just show that $0<|e_{n+1}|<\frac{1}{2+e_0}|e_n|$ and note that for $e_0>-1$ you have that $\frac{1}{2+e_0}<1$. – Shinja Feb 21 '17 at 07:16
  • I believe that only shows that $ |e_{n+1}| < |e_n|$, which doesn't imply convergence to $0$, while the argument that there is a constant factor of decrease (at least) ensures convergence to $0$ – Manuel Guillen Feb 21 '17 at 07:43
  • But once $e_0$ is chosen, $\frac{1}{2+e_0}$ is a constant (it does not depend on $n$). – Shinja Feb 21 '17 at 10:42

4 Answers4

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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 :

enter image description here

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

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 !!!.

Felix Marin
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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.
Tig la Pomme
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show with induction that for your sequence is hold $$e_n=\frac{2}{c_1 2^n+2^{n+1}-2}$$

Dr. Sonnhard Graubner
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