1. ## Investigate the sequence?

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

2. Originally Posted by khyratmath123
I have difficulty to investigate the following sequence to be convergent or divergent and find limit?
To show convergence, show it is bounded and monotonically increasing. I believe you could show $\displaystyle a_n \leq 2$ via induction.

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

3. Are you sure it's not

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

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

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

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

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

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

$\displaystyle \displaystyle L^2 = 2 + L$

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

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

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

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

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

4. thanks sir

5. The difference equation can be written as...

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

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

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

Kind regards

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