Prove the limit of this sequence exists

**AKTilted** · October 14th 2010, 01:13 PM

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:

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

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

Thanks a lot!

**Plato** · October 14th 2010, 01:43 PM

Use induction to show that $x_{n}\le x_{n+1}\le 4$ .

**chisigma** · October 14th 2010, 10:17 PM

The difference equation that defines the sequence can be written as...

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

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

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

Kind regards

$\chi$ $\sigma$

Kind regards

$\chi$ $\sigma$

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