# Prove the limit of this sequence exists

• Oct 14th 2010, 01:13 PM
AKTilted
Prove the limit of this sequence exists
Hi, I'm having a little bit trouble applying the Monotone Sequence Theorem to this question:

Determine if the sequence defined by the recursive relation:

$\displaystyle x_{k+1} = \sqrt{2+3x_k}$
$\displaystyle x_1=1$

has a limit. Present your reasoning by using the Monotone Sequence Theorem.

Thanks a lot!
• Oct 14th 2010, 01:43 PM
Plato
Use induction to show that $\displaystyle x_{n}\le x_{n+1}\le 4$.
• Oct 14th 2010, 10:17 PM
chisigma
The difference equation that defines the sequence can be written as...

$\displaystyle \displaystyle \Delta_{n} = x_{n+1}-x_{n} = \sqrt{2+3\ x_{n}}-x_{n} = f(x_{n})$ (1)

The function $\displaystyle f(*)$ is represented here...

http://digilander.libero.it/luposabatini/MHF78.bmp

There is only one 'attractive fixed point' in $\displaystyle x_{0}=3.5615528128 \dots$ and, because for any $\displaystyle x\ge -\frac{3}{2}$ is $\displaystyle |f(x)| < |x_{0}-x|$ , any $\displaystyle x_{1}\ge -\frac{3}{2}$ will produce a sequence that converges monotonically at $\displaystyle x=x_{0}$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$