Investigate the sequence?

**khyratmath123** · January 22nd 2011, 07:10 PM

I have difficulty to investigate the following sequence to be convergent or divergent and find limit?

**chiph588@** · January 22nd 2011, 07:34 PM

Originally Posted by khyratmath123

I have difficulty to investigate the following sequence to be convergent or divergent and find limit?

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To show convergence, show it is bounded and monotonically increasing. I believe you could show $a_n \leq 2$ via induction.

Once you know the limit exists (call it $L$ ), we can find it by noticing $L = \sqrt{2+L}$ .

**Prove It** · January 22nd 2011, 08:37 PM

Are you sure it's not

$\displaystyle a_1 = \sqrt{2}$ and $\displaystyle a_{n + 1} = \sqrt{2 + a_n}$ ?

If so, this is a relatively easy limit to evaluate...

It should be clear that the sequence goes $\displaystyle \left\{\sqrt{2}, \sqrt{2 + \sqrt{2}}, \sqrt{2 + \sqrt{2 + \sqrt{2}}}, \dots \right\}$ .

So you want to see if $\displaystyle \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots} }}}}$ can be evaluated (has a limit...). Call this limit $\displaystyle L$ .

If $\displaystyle L = \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots} }}}}$ then

$\displaystyle L^2 = 2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots} }}}}$

$\displaystyle L^2 = 2 + L$

$\displaystyle L^2 - L - 2 = 0$

$\displaystyle (L-2)(L+1) = 0$

$\displaystyle L - 2 = 0$ or $\displaystyle L+1 = 0$

$\displaystyle L = 2$ or $\displaystyle L = -1$ .

It should be clear that $\displaystyle L = -1$ is an extraneous solution which came from the original squaring of the equation, so the limit is $\displaystyle L = 2$ .

**khyratmath123** · January 23rd 2011, 03:32 AM

thanks sir

**chisigma** · January 23rd 2011, 08:26 AM

The difference equation can be written as...

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

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

There is only one 'attractive fixed point' in $x_{0}=2$ but that's only a necessary condition of convergence. In this case however is $|f(x)|\le |x_{0}-x|$ [red line...] so that any $a_{0}>-2$ will generate a sequence monotonically convergent at $x_{0}$ ...

Kind regards

$\chi$ $\sigma$

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