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Math Help - Infinite summation involving square root of 5

  1. #1
    Member Pranas's Avatar
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    Infinite summation involving square root of 5

    Hello.

    I have this recurring function or whatever you call it:
    \[\begin{array}{l}<br />
f(0) = 0\\<br />
f(n) = \sqrt {5 + f(n - 1)} <br />
\end{array}\]

    Once n increases, we have \[\sqrt 5 \], inside of which there is another \[\sqrt 5 \] added and so forth (this clearly converges).

    As the title says, I am looking forward to obtaining \[\mathop {\lim }\limits_{n \to \infty } f(n)\]

    P.S. I started from the greatest depth and noticed \[\sqrt {5 + \sqrt 5 }  = \frac{{\sqrt {20 + 4\sqrt 5 } }}{2} = \frac{{2 + \sqrt 5 }}{2}\], however later on I am not able to control the expression...

    P.P.S. Messing around I've got a good feeling that such infinite summations \[\sqrt {a + \sqrt {a + \sqrt {a + ...\sqrt a } } }  = \frac{{1 + \sqrt {1 + 4a} }}{2}\] can be defined like that, no clue how to prove it though
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  2. #2
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    If you can show that it converges, then:

    Take the limit of both sides of the recurrence.
    Use the fact that {\lim }\limits_{n \to \infty } f(n) = {\lim }\limits_{n \to \infty } f(n-1) = L
    to get
    L = \sqrt {5 + L}

    Solve for L.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Pranas View Post
    <br />
f(n) = \sqrt {5 + f(n - 1)} <br />
    If

    l=\displaystyle\lim_{n \to \infty}{f(n)}

    then taking limits in both menbers,

    l=\sqrt {5 + l}

    and necessarily the limit is the positive solution.


    Fernando Revilla


    Edited: Sorry, I didnīt see snowtea's post.
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  4. #4
    Member Pranas's Avatar
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    This gives \[\mathop {\lim }\limits_{n \to \infty } f(n) = \frac{{1 + \sqrt {21} }}{2}\] almost instantly.

    Smooth as silk, thanks for quick help!
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  5. #5
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    Just be aware that the above solution only works if the limit exists. To show that the limit exists, just show that the sequence is bounded above and increasing.
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    It is esay to prove that a_n=f(n)\geq 0 is increasing. To prove that is bounded we can use

    a_{n+1}^2\geq a_n^2 \Leftrightarrow\ldots \Leftrightarrow (l_1-a_n)(a_n-l_2)\geq 0

    where l_1 is the positive root of l^2-l-5 and l_2 the negative one.


    Fernando Revilla
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  7. #7
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    \displaystyle\lim_{n \to \infty}f(n)=\sqrt{5+\sqrt{5+\sqrt{5+........}}}

    \displaystyle\lim_{n \to \infty}\left[f(n)\right]^2=5+\sqrt{5+\sqrt{5+\sqrt{5+.....}}}=5+\lim_{n \to \infty}f(n)

    \displaystyle\lim_{n \to \infty}f(n)=a

    \displaystyle\ a^2-a=5\Rightarrow\ a^2-a+\frac{1}{4}=\frac{21}{4}

    \displaystyle\ \left[a-\frac{1}{2}\right]^2=\frac{21}{4}\Rightarrow\ a=\sqrt{\frac{21}{4}}+\frac{1}{2}
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