1. ## Convergence

Hi! I´m working on some problems and can´t get through this one.

Look at the following $\displaystyle [{a_{n}}]_{n=1}^{\infty}$
$\displaystyle a_{n+1} = \sqrt{2+a_n}$
where:
$\displaystyle n>1$
and:
$\displaystyle a_1 = \sqrt{2}$

Show that it converges and find the $\displaystyle lim_{n\rightarrow \infty}$

2. Originally Posted by Henryt999
Hi! I´m working on some problems and can´t get through this one.

Look at the following $\displaystyle [{a_{n}}]_{n=1}^{\infty}$
$\displaystyle a_{n+1} = \sqrt{2+a_n}$
where:
$\displaystyle n>1$
and:
$\displaystyle a_1 = \sqrt{2}$

Show that it converges and find the $\displaystyle lim_{n\rightarrow \infty}$

1) Show by induction that the sequence is monotone increasing and bounded above: $\displaystyle a_n\leq a_{n+1}\,,\,\,a_n\leq 2\,\,\forall\,n\in\mathbb{N}$ ,and deduce from that the sequence converges.

2) Now use arithmetic of limits to find the limit: if $\displaystyle \lim_{n\to\infty}a_n=\alpha$ , then $\displaystyle \alpha=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\ sqrt{2+a_n}=\sqrt{\alpha +2}$

Tonio

3. ## Thanks

Originally Posted by tonio
1) Show by induction that the sequence is monotone increasing and bounded above: $\displaystyle a_n\leq a_{n+1}\,,\,\,a_n\leq 2\,\,\forall\,n\in\mathbb{N}$ ,and deduce from that the sequence converges.

2) Now use arithmetic of limits to find the limit: if $\displaystyle \lim_{n\to\infty}a_n=\alpha$ , then $\displaystyle \alpha=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\ sqrt{2+a_n}=\sqrt{\alpha +2}$
I´m with you so far. Limit of $\displaystyle a_n$ and $\displaystyle a_{n+1}$ are the same. And how do I find $\displaystyle \alpha$? How did you get a$\displaystyle _n< 2$?

Tonio
Im new to this concept of induction.
Here is my work, is it somewhat right?
We want to prove that:
$\displaystyle a_{n+1} > a_n$

for all $\displaystyle a_n$ in N

$\displaystyle a_1 = \sqrt {2}$

$\displaystyle a_{n+1} = \sqrt{2+\sqrt{2}}$
= $\displaystyle a_{n+1} = \sqrt{2(1+2^{-0.5})}$
$\displaystyle = \sqrt2 \times [\frac {\sqrt{2}+1}{\sqrt{2}}]$
since $\displaystyle [\frac {\sqrt{2}+1}{\sqrt{2}}]$ is larger then 1 then $\displaystyle a_{n+1} < a_{n+2}$ by "induction"? Correct or something missing?

4. Originally Posted by Henryt999
Im new to this concept of induction.
Here is my work, is it somewhat right?
We want to prove that:
$\displaystyle a_{n+1} > a_n$

for all $\displaystyle a_n$ in N

$\displaystyle a_1 = \sqrt {2}$

$\displaystyle a_{n+1} = \sqrt{2+\sqrt{2}}$ <......??
= $\displaystyle a_{n+1} = \sqrt{2(1+2^{-0.5})}$
$\displaystyle = \sqrt2 \times [\frac {\sqrt{2}+1}{\sqrt{2}}]$
since $\displaystyle [\frac {\sqrt{2}+1}{\sqrt{2}}]$ is larger then 1 then $\displaystyle a_{n+1} < a_{n+2}$ by "induction"? Correct or something missing?
are you sure about that ?

5. ## Noo not at all,

No I am not and I just spotted an error you did too (I think).

$\displaystyle a_{n+1} = \sqrt{2+\sqrt{2}}$
= $\displaystyle a_{n+1} = \sqrt{2(1+2^{-0.5})}$
$\displaystyle = \sqrt2 \times [\frac {\sqrt{2}+1}{\sqrt{2}}]$
since $\displaystyle [\frac {\sqrt{2}+1}{\sqrt{2}}]$ is larger then 1 then $\displaystyle a_{n+1} < a_{n+2}$ by "induction"? Correct or something missing?[/quote] this is wrong---

It should be: $\displaystyle a_{n+1} = \sqrt{2+\sqrt{2}}$
$\displaystyle = \sqrt{2(1+\frac{1}{\sqrt{2}})}$
$\displaystyle = \sqrt{2} \times \sqrt{1+\frac{1}{\sqrt{2}}}$
And since $\displaystyle \sqrt{1+\frac{1}{\sqrt{2}}}$ can only be larger then 1 by induction $\displaystyle a_{n+1} >a_n$ ?? (I think)

6. you're quoting yourself ,i didn't write anything,"Tonio" told you exactly what must be done.
anyway,$\displaystyle a_{n+1} = \sqrt{2+\sqrt{2}}$ doesn't mean anything .
it's $\displaystyle a_{2} = \sqrt{2+\sqrt{2}}$

7. ## Yes

Well, I thank tonio and you for your help, but Iám sorry I do not fully understand. That "quote" wasnt suppose to be there.
$\displaystyle a_2 >a_1$
So far so god.
Now I have to show that the upper bound is 2, and that is where I fail.
How do I know the upperbound?
Tonio did $\displaystyle lim_{n\rightarrow \infty} = \sqrt{2+ \alpha}$

How do I solve find this $\displaystyle \alpha$ It must be 2, but I cant prove that..
By induction all $\displaystyle a_{n}>a_{n-1}$ and the $\displaystyle lim [a_n] = \sqrt{2+2}$ but I dont know how to prove it.

You dont mean that anyone is suppose to do the exercise for me, but I would prefer explanation by baby steps...

8. well the exercise should mention that the upper bound is $\displaystyle 2$.
and all you have to do is prove it.
anyway,try to prove $\displaystyle a_n\leq 2,\forall n\in \mathbb{N}$.
How ?
it's right for $\displaystyle n=1$ we have $\displaystyle \sqrt{2}\leq 2$,suppose you have $\displaystyle a_n\leq 2$ for some rank $\displaystyle n$ and try to prove $\displaystyle a_{n+1}\leq 2$ for some rank $\displaystyle n+1$.

9. if you succeeded in proving $\displaystyle a_n\leq 2,\forall n\in \mathbb{N}$
what can you say about $\displaystyle a_{n+1}-a_n$.

10. ## Like this?

$\displaystyle a_{2}>a_{1}$
because $\displaystyle \sqrt{2}<2$

$\displaystyle a_3>a_2$
because
$\displaystyle 2>$$\displaystyle \sqrt{2 +\sqrt {2}} for \displaystyle a_4>a_{3} \displaystyle 2>\sqrt{2+\sqrt{2 +\sqrt{2}}} Therefor it is true for all \displaystyle n \in N Now I want to find where it converges too. Since the \displaystyle \lim [a_{n+1}] = \lim [a_n] we have \displaystyle a = \sqrt{2+a} then \displaystyle a^2 = 2+a then \displaystyle a^2 -a-2=0 then \displaystyle a = 0.5 \pm \sqrt{\frac{1}{2}+2} \displaystyle a = 0.5 \pm 1.5 \displaystyle a_1 = -1 and \displaystyle a_2 = 2 Since a can not be negative cause\displaystyle \sqrt{2} <a_n<a_{n+1}\leq {2} Then it must converge towards \displaystyle \sqrt {2+2} = 2 Does this make sence or is my prof going to \displaystyle \Rightarrow 11. ## If then........ Originally Posted by Raoh if you succeeded in proving \displaystyle a_n\leq 2,\forall n\in \mathbb{N} what can you say about \displaystyle a_{n+1}-a_n. If I succed then \displaystyle 2-\sqrt{2}>a_{n+1}-a_n >0 12. Originally Posted by Henryt999 \displaystyle a_{2}>a_{1} because \displaystyle \sqrt{2}<2 \displaystyle a_3>a_2 because \displaystyle 2>$$\displaystyle \sqrt{2 +\sqrt {2}}$

for $\displaystyle a_4>a_{3}$
$\displaystyle 2>\sqrt{2+\sqrt{2 +\sqrt{2}}}$
Therefor it is true for all $\displaystyle n \in N$ <..... you still need to check until GOD KNOWS WHERE.

Now I want to find where it converges too.
Since the $\displaystyle \lim [a_{n+1}] = \lim [a_n]$=some real number <.... we can't say that until we are sure that the limit exists.
we have

$\displaystyle a = \sqrt{2+a}$
then
$\displaystyle a^2 = 2+a$
then
$\displaystyle a^2 -a-2=0$
then
$\displaystyle a = 0.5 \pm \sqrt{\frac{1}{2}+2}$
$\displaystyle a = 0.5 \pm 1.5$

$\displaystyle a_1 = -1$
and
$\displaystyle a_2 = 2$
Since a can not be negative cause$\displaystyle \sqrt{2} <a_n<a_{n+1}\leq {2}$
Then it must converge towards $\displaystyle \sqrt {2+2} = 2$
Does this make sence or is my prof going to \Rightarrow
you agree that $\displaystyle a_n\leq 2,\forall n\in \mathbb{N}$ (assuming that you proved it by induction)
$\displaystyle a_{n+1}-a_{n}=\sqrt{2+a_n}-a_n=\frac{2+a_n-a_n^2}{\sqrt{2+a_n}+a_n}\geq 0,\forall a_n\leq 2$

13. Originally Posted by Henryt999
Hi! I´m working on some problems and can´t get through this one.

Look at the following $\displaystyle [{a_{n}}]_{n=1}^{\infty}$
$\displaystyle a_{n+1} = \sqrt{2+a_n}$
where:
$\displaystyle n>1$
and:
$\displaystyle a_1 = \sqrt{2}$

Show that it converges and find the $\displaystyle lim_{n\rightarrow \infty}$
Henryt999 ,first of all you must write down a few terms of the sequence to get a "feeling" of what the sequence look like.

And we have that:

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

$\displaystyle a_{2} = \sqrt{2+\sqrt{2}}$

$\displaystyle a_{3} = \sqrt{2+\sqrt{2+\sqrt{2}}}$.

From the above we have that: $\displaystyle 0<a_{1}<a_{2}<a_{3}$.

So we see that the sequence is positive and increasing ,hence can we prove in general that :

1)$\displaystyle a_{n}>0$?

2) $\displaystyle \frac {a_{n}}{a_{n+1}}<1$?

The first one we can prove by induction .

For the 2nd one we must prove: $\displaystyle \frac{a_{n}}{\sqrt{2+a_{n}}}<1$ ,or $\displaystyle \frac{(a_{n})^2}{2+a_{n}}<1$,since we have proved that $\displaystyle a_{n}>0$.

...............or................................. ...........

$\displaystyle (a_{n}-2)(a_{n}+1)<0$.

For that you must prove that: $\displaystyle a_{n}<2$ for all nεN ,since $\displaystyle a_{n}+1>0$ for all nεN.

So far having proved that the sequence is psitive increasing and bounded from above by 2 ,the sequence definitely converges to a limit x>0

Now if a sequence converges to x ,every subsequence of that sequence converges to x ,hence :

$\displaystyle lim_{n\to\infty}a_{n} = lim_{n\to\infty}a_{n+1} = x$,since $\displaystyle (a_{n+1})$ is a subsequence of $\displaystyle a_{n}$.

Thus $\displaystyle lim_{n\to\infty}a_{n+1}= \sqrt{2+lim_{n\to\infty}a_{n}}\Longrightarrow x=\sqrt{2+x}$.

AND x=-1 or x=2

14. ## Yes thank you

I´m trying to prove it by induction that $\displaystyle a_n<a_{n+1}$

$\displaystyle a_1 = \sqrt{2}$
$\displaystyle a_{2} = \sqrt{2+\sqrt{2}}$
hence $\displaystyle a_2>a_1$

and $\displaystyle a_3 = \sqrt{2+\sqrt{2+\sqrt{2}}}$

$\displaystyle a_3>a_2$
$\displaystyle a_1<a_2<a_3_<a_4$
And then it follows by "induction magic" that for all $\displaystyle a_{n+1}>a_n$
Is this some proof?

15. So we see that the sequence is positive and increasing ,hence can we prove in general that :

1)$\displaystyle a_{n}>0$?

2) $\displaystyle \frac {a_{n}}{a_{n+1}}<1$?

The first one we can prove by induction .

For the 2nd one we must prove: $\displaystyle \frac{a_{n}}{\sqrt{2+a_{n}}}<1$ ,or $\displaystyle \frac{(a_{n})^2}{2+a_{n}}<1$,since we have proved that $\displaystyle a_{n}>0$.
I feel that your second statement is much too complicated. It is possible to show that $\displaystyle a_{n-1}<a_{n}$for all $\displaystyle n$ via induction, as well.

You have already checked some base cases, so suppose that $\displaystyle a_{k-1}<a_k$. Now, add two to each side. $\displaystyle a_{k-1}+2<a_k+2$. Now take the square root of both sides, keeping in mind that these terms are positive and square root is monotonic. $\displaystyle \sqrt{a_{k-1}+2}<\sqrt{a_k+2}$. But, by definition, $\displaystyle \sqrt{a_{k-1}+2}=a_k$ and $\displaystyle \sqrt{a_k+2}=a_{k+1}$. So we have $\displaystyle a_k<a_{k+1}$. This fulfills the induction step and hence, monotonicity is proved.

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