Proof of Convergence for Recursive Sequence

**moemoe** · June 3rd 2010, 07:18 AM

I am trying to figure out how to prove that the following sequence converges:

$a_{n+1}=\sqrt{2+\sqrt{a_{n}}}$

with $a_{1}=\sqrt{2}$

Thoughts anyone?

**moemoe** · June 3rd 2010, 07:53 AM

Originally Posted by General

Since the sequence converges, then $\lim_{n\to\infty} a_n = L$ , where L is a finite number ..

Recall that, for the convergent sequence $a_n$ :
$\lim_{n\to\infty} a_n=\lim_{n\to\infty} a_{n+1}$ ..

Since : $a_{n+1}=\sqrt{2+\sqrt{a_{n}}}$

By taking the limit of both sides, we will conclude the following equation :

$L=\sqrt{2+\sqrt{L}}$ , solve it for L ..

Thanks for the quick replies. I believe that in a formal proof

$L=\sqrt{2+\sqrt{L}}$

can be used to solve for the limit L if we know that the series converges. However, I need to first prove that the series converges given

$a_{1}=\sqrt{2}$

**parkhid** · June 3rd 2010, 07:54 AM

I think we should say this :

$|a_{n+1} - L|<\epsilon$

and $-\epsilon < a_{n+1} - L < \epsilon$

it means :

$0 < a_{n+1} - L < 2\epsilon$

and $a_{n+1} < 2\epsilon+L$

also if $2\epsilon<<L$ we can unspot $2\epsilon$ .
and $a_{n+1} < L$

this is the proof. but to complete it

we should proof that $2\epsilon<<L$ . or

$\epsilon<<\frac{L}{2}$

We Know L is 1 as here :

$\lim_{n\rightarrow\infty}\sqrt{\frac{2}{a_{n}} +1} = 1$

so the $\epsilon<<0.5$

**Plato** · June 3rd 2010, 07:57 AM

Originally Posted by moemoe

However, I need to first prove that the series converges given

Show that the sequence is increasing and bounded.
Use induction to do that.

**Defunkt** · June 3rd 2010, 08:00 AM

Originally Posted by parkhid

I think we should say this :

$|a_{n+1} - L|<\epsilon$
...

You can't assume convergence if you're trying to prove it...

@moemoe: The usual way to prove that this type of recursive sequences converge is to prove by induction that the sequence is monotone and bounded.

In your case, plug in the values for the first few terms to see whether the sequence is increasing or decreasing, and then prove it by induction. After you do that, prove that it is bounded (this can be done by induction as well).

EDIT: Plato ahead of me :<

**moemoe** · June 3rd 2010, 03:53 PM

So assuming

$ a_{n} \geq a_{n-1} \geq 0 $

we have

$ a_{n+1} = \sqrt{2+\sqrt{a_{n}}} \geq \sqrt{2+\sqrt{a_{n-1}}}= a_{n} $

Does anyone have a slick way of showing the sequence is bounded?

**moemoe** · June 3rd 2010, 03:59 PM

Maybe define a new sequence

$ b_{n+1}=\sqrt{2+\sqrt{b_{n}}} $ with $ b_{1}=2 $

where $ a_{n} \leq b_{n} $ for each $n$ and $ b_{n} $ is decreasing by the same induction argument above?

Is there another simpler way?

**Defunkt** · June 3rd 2010, 04:08 PM

There is a simpler way.

First, validate that $a_1, a_2 < 2$ .

Now, assume $a_n < 2$ and use the induction hypothesis to prove that $a_{n+1} < 2$ .
This will prove that the sequence is bounded by 2 (it doesn't mean that 2 is the limit, though..)

**galactus** · June 3rd 2010, 05:47 PM

Perhaps we can try this.

By the induction step it can be shown that $a_{n}<2$

$a_{n+1}=\sqrt{2+\sqrt{a_{n}}}$

$(a_{n+1})^{2}=2+\sqrt{a_{n}}$

$(a_{n+1})^{2}-1=1+\sqrt{a_{n}}$

$(a_{n+1}+1)(a_{n+1}-1)=1+\sqrt{a_{n}}$

Since for all n, $a_{n}<2$

$a_{n+1}-1<1$

$a_{n+1}+1>a_{n}+1$

$a_{n+1}>a_{n}$

and it is monotone increasing.

**chisigma** · June 4th 2010, 01:59 AM

The sequence is defined recursively as...

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

The function $f(x)= \sqrt{2 + \sqrt{x}} - x$ is represented here...

It has a single zero at $x_{0} = 1.83117720721...$ and because $|f(x)|$ is less that the line crossing the x axis in $x=x_{0}$ with unity negative slope, any $a_{0} \ge 0$ will produce a sequence converging at $x_{0}$ without oscillations...

Kind regards

$\chi$ $\sigma$

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