I am trying to figure out how to prove that the following sequence converges:
$\displaystyle a_{n+1}=\sqrt{2+\sqrt{a_{n}}}$
with $\displaystyle a_{1}=\sqrt{2}$
Thoughts anyone?
I think we should say this :
$\displaystyle |a_{n+1} - L|<\epsilon$
and $\displaystyle -\epsilon < a_{n+1} - L < \epsilon $
it means :
$\displaystyle 0 < a_{n+1} - L < 2\epsilon $
and $\displaystyle a_{n+1} < 2\epsilon+L$
also if $\displaystyle 2\epsilon<<L$ we can unspot $\displaystyle 2\epsilon$ .
and $\displaystyle a_{n+1} < L$
this is the proof. but to complete it
we should proof that $\displaystyle 2\epsilon<<L$ . or
$\displaystyle \epsilon<<\frac{L}{2}$
We Know L is 1 as here :
$\displaystyle \lim_{n\rightarrow\infty}\sqrt{\frac{2}{a_{n}} +1} = 1$
so the $\displaystyle \epsilon<<0.5$
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 :<
Maybe define a new sequence
$\displaystyle
b_{n+1}=\sqrt{2+\sqrt{b_{n}}}
$ with $\displaystyle
b_{1}=2
$
where $\displaystyle
a_{n} \leq b_{n}
$ for each $\displaystyle n$ and $\displaystyle
b_{n}
$ is decreasing by the same induction argument above?
Is there another simpler way?
There is a simpler way.
First, validate that $\displaystyle a_1, a_2 < 2$.
Now, assume $\displaystyle a_n < 2$ and use the induction hypothesis to prove that $\displaystyle a_{n+1} < 2$.
This will prove that the sequence is bounded by 2 (it doesn't mean that 2 is the limit, though..)
Perhaps we can try this.
By the induction step it can be shown that $\displaystyle a_{n}<2$
$\displaystyle a_{n+1}=\sqrt{2+\sqrt{a_{n}}}$
$\displaystyle (a_{n+1})^{2}=2+\sqrt{a_{n}}$
$\displaystyle (a_{n+1})^{2}-1=1+\sqrt{a_{n}}$
$\displaystyle (a_{n+1}+1)(a_{n+1}-1)=1+\sqrt{a_{n}}$
Since for all n, $\displaystyle a_{n}<2$
$\displaystyle a_{n+1}-1<1$
$\displaystyle a_{n+1}+1>a_{n}+1$
$\displaystyle a_{n+1}>a_{n}$
and it is monotone increasing.
The sequence is defined recursively as...
$\displaystyle \Delta_{n} = a_{n+1} - a_{n} = \sqrt{2 + \sqrt{a_{n}}} - a_{n} = f(a_{n})$ (1)
The function $\displaystyle f(x)= \sqrt{2 + \sqrt{x}} - x$ is represented here...
It has a single zero at $\displaystyle x_{0} = 1.83117720721...$ and because $\displaystyle |f(x)|$ is less that the line crossing the x axis in $\displaystyle x=x_{0}$ with unity negative slope, any $\displaystyle a_{0} \ge 0$ will produce a sequence converging at $\displaystyle x_{0}$ without oscillations...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$