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Thread: Convergence

  1. #1
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    Convergence

    Hi! Im working on some problems and cant 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 $\displaystyle lim_{n\rightarrow \infty}$
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  2. #2
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    Quote Originally Posted by Henryt999 View Post
    Hi! Im working on some problems and cant 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
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  3. #3
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    Quote Originally Posted by tonio View Post
    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 $\displaystyle \alpha=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\ sqrt{2+a_n}=\sqrt{\alpha +2}$
    Im 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?
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  4. #4
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    Smile

    Quote Originally Posted by Henryt999 View Post
    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 ?
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  5. #5
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    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)
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    Smile

    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 $\displaystyle a_{2} = \sqrt{2+\sqrt{2}}$
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  7. #7
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    Yes

    Well, I thank tonio and you for your help, but Im 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...
    Last edited by Henryt999; Feb 11th 2010 at 05:34 AM.
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    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 $\displaystyle n+1$.
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    if you succeeded in proving $\displaystyle a_n\leq 2,\forall n\in \mathbb{N}$
    what can you say about $\displaystyle a_{n+1}-a_n$.
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  10. #10
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    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 $\displaystyle \Rightarrow$
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  11. #11
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    If then........

    Quote Originally Posted by Raoh View Post
    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 $\displaystyle 2-\sqrt{2}>a_{n+1}-a_n >0$
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  12. #12
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    Quote Originally Posted by Henryt999 View Post
    $\displaystyle a_{2}>a_{1}$
    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)
    $\displaystyle a_{n+1}-a_{n}=\sqrt{2+a_n}-a_n=\frac{2+a_n-a_n^2}{\sqrt{2+a_n}+a_n}\geq 0,\forall a_n\leq 2$
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  13. #13
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    Quote Originally Posted by Henryt999 View Post
    Hi! Im working on some problems and cant 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
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  14. #14
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    Yes thank you

    Im 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?
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  15. #15
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    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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