Converge(not easy)

silviu

Having $(x_n)_{n>0}$ with $x_1,x_2,x_3>0$ and $x_{n + 3} = \frac{{x_{n + 2}^2 + 5x_{n + 1}^2 + x_n^2 }}{{x_{n + 2} + 5x_{n + 1} + x_n }}\forall n \ge 1$
Prove that $(x_n)_{n>0}$ converge.

**Posted:** Sat Feb 19, 2005 11:58 am

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Love for math

fleeting_guest · Yang-Mills Theory Offline Joined: 14 Dec 2004 Posts: 900

**Posted:** Sun Feb 20, 2005 7:50 pm

fleeting_guest · Yang-Mills Theory Offline Joined: 14 Dec 2004 Posts: 900

A better (more constructive) version of the same argument:
there exists some positive constant $c < 1$ so that
$|I_{n+1}| \leq c|I_n|$ .

**Posted:** Mon Feb 21, 2005 4:05 pm

the game

yes fleeting guest its called contractive sequences.They are Cauchy sequence & are hence convergent(R is complete)

**Posted:** Tue Feb 22, 2005 12:14 am

the game

O sorry I am horribly wrong.The contractive sequences are those:

mod(xn+1 -xn) <= c*mod(xn-xn-1) where 0<c<1

Sorry for the mistake..

**Posted:** Tue Feb 22, 2005 11:19 am

fleeting_guest · Yang-Mills Theory Offline Joined: 14 Dec 2004 Posts: 900

The argument above does not show that $d_n = |x_n - x_{n-1}|$ is contracting, it shows $d_{n+3} < c d_n$ . This certainly implies convergence, but in the above argument it is enough that $|I_n| \to 0$ (not necessarily geometrically). It would be interesting to prove directly that $d_{n+1} < c d_n$ , if that is true.

**Posted:** Tue Feb 22, 2005 1:34 pm