6

Show that the sequence $(x_n)$ defined as $x_{n+1}=\frac{1}{1+x_n},x_0=0$ converge. My attempt is to show that $(x_n)$ is Cauchy so i would like to have a feedback on my proof,please.

First of all we show by induction that $2>x_n\ge0$. The previous inequality is true for $n=0$ so we can suppose that it works up to a certain $n$. We want to proof right now that it holds fot $n+1$ and so show that $2>x_{n+1}\ge0$:

$2>x_{n+1}\ge0 \iff 2>\frac{1}{1+x_n}\ge0$.

It is trivial that $x_{n+1}\ge0$ by induction hypothesis (because $x_n\ge0$). Moreover, as $2>x_n\ge0$, we have that

$x_{n+1}=\frac{1}{1+x_n}\le1<2$. So we can conclude that $2>x_n\ge0 \ \forall n\ge0$

Then, $\forall p\ge2$ we have:

$|x_p-x_{p-1}|=\Big|\frac{x_{p-2}-x_{p-1}}{(1+x_{p-1})(1+x_{p-2})}\Big|\le 1 \cdot|x_{p-2}-x_{p-1}|=...=1^{p-1}|x_0-x_1|=1^{p-1}$

We show now that $(x_n)$ is Cauchy. So, by définition of Cauchy sequence, we have to show that $\forall \epsilon>0 \ \exists N \ \forall m,n>N$: $|x_m-x_n|<\epsilon$. But,

$|x_m-x_n|\le|x_m-x_{m-1}|+|x_{m-1}+x_{m-2}|+...+|x_{n+1}-x_n|\le(m-n-1)$

So $(x_n)$ is Cauchy

I am not reall sure if this proof holds, because i think that in case where $m=n-1$ it doesn't work,or...?

Daniil
  • 1,647
  • 1
    You can simply prove $(x_n)_n$ is monotonically decreasing (by induction) on your interval. Since its' bounded and monotonic, it's convergent and every convergent sequence in $\Bbb R$ is Cauchy. – PinkyWay Feb 13 '21 at 17:17
  • 3
    What's wrong with your proof is not just that it doesn't work when $m = n - 1$, but it also fails to make $|x_m - x_n|$ small. A proof that $(x_n)$ is Cauchy should be able to tell me when $|x_m - x_n| < \frac{1}{2}$, for example, which your proof does not. – Theo Bendit Feb 13 '21 at 17:20
  • @Invisible oh well seen! Thank you for yor answer – Daniil Feb 13 '21 at 17:21
  • 1
    You actually haven't proven the statement about the Cauchy sequence because $\varepsilon$ should be arbitrary and $x_p-x_{p-1}\ne0$. Your $m$ and $n$ are not fixed, so I fail to see how you've proven the sequence is bounded. – PinkyWay Feb 13 '21 at 17:25
  • @Invisible Yes, im agree with that. I found it strange as well when i bounded $|x_p-x_{p-1}|$ by $1$.. But, i don't really see where it fails as i don't understand why $|x_p-x_{p-1}|=\Big|\frac{x_{p-2}-x_{p-1}}{(1+x_{p-1})(1+x_{p-2})}\Big|\le|x_{p-2}-x_{p-1}|$ would be false – Daniil Feb 13 '21 at 17:27
  • How did you get $|x_p-x_{p-1}|=\cdots=1^{p-1}|x_0-x_1|$? – PinkyWay Feb 13 '21 at 17:32
  • See this. – PinkyWay Feb 13 '21 at 17:36
  • @Invisible $|x_p-x_{p-1}|=|\frac{x_{p-2}-x_{p-1}}{(1+x_{p-1})(1+x_{p-2})}|$. Then, as $2>x_n\ge0$, i bounded the denominator by $(1+0)(1+0)$ – Daniil Feb 13 '21 at 17:37
  • You cannot put the equality whatsoever. Also see this. Revise the monotone convergence theorem. (=: – PinkyWay Feb 13 '21 at 17:38
  • @Invisible alright, thank you. I will check your links – Daniil Feb 13 '21 at 17:40
  • Here is a proof of a more general case. – rtybase Feb 14 '21 at 13:10
  • Just for completeness the explicit solution of the recursion: letting $x(n) = p(n)/q(n)$ the recursion results in two recursions $p(n+1)=q(n), q(n+1)=q(n)+p(n)$ from which follows $p(n+2) = p(n+1)+p(n)$ which with $p(0)=0,p(1)=1$ is solved by the Fibonacci numbers, and $q(n+2) = q(n+1)+q(n)$ which with $q(0)=1,q(1)=1$ is solved by $\frac{1}{2}(-\text{Fibonacci}(n) + \text{Lucas}(n))$ where the Lucas numbers appear. The quotient of $p(n)/q(n)$ approaches the golden ratio as $n\to\infty$ – Dr. Wolfgang Hintze Nov 12 '21 at 15:36

1 Answers1

4

Let $x_{1}>0$ and $\displaystyle x_{n+1}=\frac{1}{1+x_{n}}$. Notice that $x_{n}>0$ (by induction) and that : $$ |x_{n+2}-x_{n+1}|=\left|\frac{1}{1+x_{n+1}}-\frac{1}{1+x_{n}}\right|=\frac{|x_{n}-x_{n+1}|}{(1+x_{n+1})(1+x_{n})}\leq\frac{4}{9}|x_{n}-x_{n+1}| $$ Hence, $x_{n}$ is a contraction and hence it is Cauchy, and so converges by completeness of $(\mathbb{R},|.|)$. Let $L$ be its limit. We have that $\displaystyle L=\frac{1}{1+L}$ that is : \begin{align*} 0&=L^{2}+L-1\\&=(L+0.5)^{2}-0.25-1\\&=\left(L-\frac{-1+\sqrt{5}}{2}\right)\left(L-\frac{-1-\sqrt{5}}{2}\right) \end{align*} We obtain that either $\displaystyle L=\frac{-1+\sqrt{5}}{2}$, or $\displaystyle L=\frac{-1-\sqrt{5}}{2}$. However, since $L\geq0$ and since $x_{n}>0$, we conclude that : $$ \lim_{n\to\infty} x_{n}=\frac{-1+\sqrt{5}}{2} $$ where $\displaystyle\varphi:=\frac{-1+\sqrt{5}}{2}$ is the golden ratio.

As $\text{Theo Bendit}$ mentioned, What's wrong with your proof is that it fails to make $\left|x_{m}-x_{n}\right|$ small.

B E I R U T
  • 1,818
  • Thank you for an answer, but i still want to understand what is going wrong in my proof. – Daniil Feb 13 '21 at 17:20
  • I just have one question: how did you find out that $\frac{|x_{n}-x_{n+1}|}{(1+x_{n+1})(1+x_{n})}\leq\frac{1}{4}|x_{n}-x_{n+1}|$? I mean from where di you get that $1/4$? My question might be stupid but still.. – Daniil Feb 13 '21 at 18:48
  • 1
    First of all, I removed the absolute because the denominator is positive and it can be proven by induction. Second, $1+x_{n+1}>1$ and $1+x_{n}>1$ why? because $x_{n}$ and $x_{n+1}$ are positive. This would mean that by multiplying them together and then reciprocating them in the denominator means its less that $\frac{1}{2}$ so I chose $\frac{1}{4}$ by mistake but is still correct. I will fix it anyways. – B E I R U T Feb 13 '21 at 18:53
  • 1
    Alright, thank you! – Daniil Feb 13 '21 at 19:03
  • Anytime $\text{:-)}$ – B E I R U T Feb 13 '21 at 19:03
  • Okey , one more question xD Are we agree that it would be legal to take for example $3/2$ and not $1/2$? Simply it is not gonna work as it won't make $|x_m-x_n|$ small enough – Daniil Feb 13 '21 at 19:09
  • 2
    Hm... But why $5/2$ would work? If you take $m=n+1$ (notations in my proof), and $\epsilon=3$ it won't work, or? – Daniil Feb 13 '21 at 19:17
  • by mistake my bad $:)$ – B E I R U T Feb 13 '21 at 19:18
  • 2
    It looks like there is a typographical error. The inequality $|x_{n+2}-x_{n+1}| \leqslant \frac{3}{2} | x_n-x_{n+1}|$ appears too weak for the conclusion. But as it happens it is not hard to see $1 \geqslant x_n \geqslant \frac{1}{2}$ for all $n \geqslant 1$ leading to $|X_{n+2}-x_{n+1}| \leqslant \frac{4}{9}| x_n-x_{n+1}|$, which is sufficient. – WA Don Feb 14 '21 at 10:27