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Math Help - Convergence

  1. #1
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    Convergence

    Hi! Im working on some problems and cant get through this one.

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

    Show that it converges and find the 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 [{a_{n}}]_{n=1}^{\infty}
     a_{n+1} = \sqrt{2+a_n}
    where:
    n>1
    and:
    a_1 = \sqrt{2}

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

    1) Show by induction that the sequence is monotone increasing and bounded above: 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 \lim_{n\to\infty}a_n=\alpha , then \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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    Thanks

    Quote Originally Posted by tonio View Post
    1) Show by induction that the sequence is monotone increasing and bounded above: 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 \lim_{n\to\infty}a_n=\alpha , then \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 a_n and a_{n+1} are the same. And how do I find \alpha? How did you get a _n< 2?

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

    for all a_n in N

    a_1 = \sqrt {2}

    a_{n+1} = \sqrt{2+\sqrt{2}}
    = a_{n+1} = \sqrt{2(1+2^{-0.5})}
    = \sqrt2 \times [\frac {\sqrt{2}+1}{\sqrt{2}}]
    since [\frac {\sqrt{2}+1}{\sqrt{2}}] is larger then 1 then 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:
    a_{n+1} > a_n

    for all a_n in N

    a_1 = \sqrt {2}

    a_{n+1} = \sqrt{2+\sqrt{2}} <......??
    = a_{n+1} = \sqrt{2(1+2^{-0.5})}
    = \sqrt2 \times [\frac {\sqrt{2}+1}{\sqrt{2}}]
    since [\frac {\sqrt{2}+1}{\sqrt{2}}] is larger then 1 then 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).


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

    It should be: a_{n+1} = \sqrt{2+\sqrt{2}}
    = \sqrt{2(1+\frac{1}{\sqrt{2}})}
    = \sqrt{2} \times \sqrt{1+\frac{1}{\sqrt{2}}}
    And since \sqrt{1+\frac{1}{\sqrt{2}}} can only be larger then 1 by induction a_{n+1} >a_n ?? (I think)
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  6. #6
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    Smile

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

    How do I solve find this \alpha It must be 2, but I cant prove that..
    By induction all a_{n}>a_{n-1} and the 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; February 11th 2010 at 05:34 AM.
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  8. #8
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    well the exercise should mention that the upper bound is 2.
    and all you have to do is prove it.
    anyway,try to prove a_n\leq 2,\forall n\in \mathbb{N}.
    How ?
    it's right for n=1 we have \sqrt{2}\leq 2,suppose you have a_n\leq 2 for some rank n and try to prove a_{n+1}\leq 2 for some rank n+1.
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  9. #9
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    Smile

    if you succeeded in proving a_n\leq 2,\forall n\in \mathbb{N}
    what can you say about a_{n+1}-a_n.
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  10. #10
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    Like this?

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

    a_3>a_2
    because
    2> \sqrt{2 +\sqrt {2}}

    for a_4>a_{3}
    2>\sqrt{2+\sqrt{2 +\sqrt{2}}}
    Therefor it is true for all n \in N

    Now I want to find where it converges too.
    Since the \lim [a_{n+1}] = \lim [a_n] we have

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

    a = 0.5 \pm 1.5

    a_1 = -1
    and
    a_2 = 2
    Since a can not be negative cause  \sqrt{2} <a_n<a_{n+1}\leq {2}
    Then it must converge towards \sqrt {2+2} = 2
    Does this make sence or is my prof going to \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 a_n\leq 2,\forall n\in \mathbb{N}
    what can you say about a_{n+1}-a_n.
    If I succed then 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
    a_{2}>a_{1}
    because \sqrt{2}<2

    a_3>a_2
    because
    2> \sqrt{2 +\sqrt {2}}

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

    Now I want to find where it converges too.
    Since the \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

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

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

    Show that it converges and find the 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:

    a_{1} = \sqrt{2}

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

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

    From the above we have that: 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) a_{n}>0?

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

    The first one we can prove by induction .

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

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


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

    For that you must prove that: a_{n}<2 for all nεN ,since 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 :

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

    Thus  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 a_n<a_{n+1}

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

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

    a_3>a_2
    a_1<a_2<a_3_<a_4
    And then it follows by "induction magic" that for all 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) a_{n}>0?

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

    The first one we can prove by induction .

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

    You have already checked some base cases, so suppose that a_{k-1}<a_k. Now, add two to each side. 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. \sqrt{a_{k-1}+2}<\sqrt{a_k+2}. But, by definition, \sqrt{a_{k-1}+2}=a_k and \sqrt{a_k+2}=a_{k+1}. So we have a_k<a_{k+1}. This fulfills the induction step and hence, monotonicity is proved.
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