Let sequence $\{x_n\}$ is defined following: $$x_1=2,\quad x_2=2+\dfrac{1}{2},\quad x_3=2+\dfrac{1}{2 + \dfrac{1}{2}}, \cdots$$

Evaluate $\lim \limits_{n\to \infty}x_n$.

My sketch proof: I thought that $\{x_{2n-1}\}$ increases but $\{x_{2n}\}$ decreases. For any $n$ we have $2\leqslant x_n<3$. In that case both subsequences converges to same limit hence $\{x_n\}$ also converges to that limit.

The main question is how to prove that $\{x_{2n-1}\}$ increases and $\{x_{2n}\}$ decreases. Can anyone help to me please?

P.S. Please do not diplicate this topic with this because here was using Banach fixed-point theorem.

  • $\begingroup$ Well, you have to use recursion and induction a lot. I tried to do that, but it took me a lot of ink and paper. After the original question has been closed, I gave up. $\endgroup$ – Crostul Oct 28 '15 at 14:35
  • $\begingroup$ @Crostul, sorry please but you mean to use induction? $\endgroup$ – ZFR Oct 28 '15 at 14:38
  • 1
    $\begingroup$ Look at the wiki page on continued fractions, esp the theorems 1-5 on the properties satisfied by the convergents. $\endgroup$ – Aravind Oct 28 '15 at 14:39

Hint: If $x_n<x_{n+2}$ then $$x_{n+1}=2+\frac1{x_n}>2+\frac1{x_{n+2}}=x_{n+3}$$ and in the same way $x_{n+2}<x_{n+4}$.

In the same way, one finds for index difference $1$ if $x_n<x_{n+1}$ then $x_{n+1}>x_{n+2}$.

In short, it is easy to prove your assertions with induction, using $x_1<x_2$ and $x_1<x_3$ as induction start to show

  • $\{x_{2m-1}\}_{m\in\Bbb N}$ increasing,
  • $\{x_{2m}\}_{m\in\Bbb N}$ decreasing and
  • $x_{2k-1}<x_{2n-1}<x_{2n}<x_{2m}$ for all $k,m$ and some $n\ge\max(k,m)$
  • $\begingroup$ Dear LutzL! Are you using here math induction? $\endgroup$ – ZFR Oct 28 '15 at 16:06
  • $\begingroup$ Yes, using $x_1<x_2$ and $x_1<x_3$ to show $x_{2m-1}$ increasing, $x_{2m}$ decreasing and $x_{2k-1}<x_{2n-1}<x_{2n}<x_{2m}$ for some $n\ge\max(k,m)$. $\endgroup$ – LutzL Oct 28 '15 at 16:24
  • $\begingroup$ Really nice solution! Thanks a lot, dear LutzL! :) $\endgroup$ – ZFR Oct 28 '15 at 16:32

Hint: Write $x_n = \tfrac{a_n}{b_n}$. Then we have a sequence $(a_n,b_n)$ of natural numbers satisfying $$\pmatrix{a_{n+1}\\b_{n+1}} = \pmatrix{2&1\\1&0}\pmatrix{a_n\\b_n}.$$ With just a bit of linear algebra, this allows you to find an explicit form for $x_n$ as a fraction. If you can do this, then you'll be easily able to conclude.

  • $\begingroup$ For this we must find $n$th power of $2\times 2$ matrix right? $\endgroup$ – ZFR Oct 28 '15 at 14:50
  • $\begingroup$ @RFZ Exactly.${}$ $\endgroup$ – Daniel Robert-Nicoud Oct 28 '15 at 18:17

We have $x_{n+1}=2+1/x_n.$ Assuming the limit $L$ exists ,we have $L\geq 2$ and $L=2+1/L.$This is valid because if we put $$a_n=L+d_n$$ then $$L+d_{n+1}=2+1/(L+d_n)$$ which cannot hold for non-zero $L$ and for values $d_{n+1},d_n$ arbitrarily close to $0$ unless $L=2+1/L$....Now $$L>0\wedge L=2+1/L\iff L>0\wedge L^2=2L+1\iff L=1+\sqrt 2.$$ To see that $a_n$ actually does converge to $1+\sqrt 2$, put $$a_n=1+\sqrt 2+d_n=L+d_n.$$ $$\text {Then } L+d_{n+1}=2+1/(L+d_n)$$ which (using the simplifying formula $L^2=2L+1$ ) implies $$|d_{n+1}|= |d_n(2-L)/(2+d_n)|=|d_n(2-L)/a_n|\leq |d_n(2-L)/2|=|d_n|(\sqrt 2-1)/2.$$ Since $0<(\sqrt 2-1)/2<1$,we have $\lim_{n\to \infty}d_n=0$.Observe how exploring the consequences of an assertion can help to prove or disprove it: "Assuming the limit exists...."


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.