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Math Help - limit of a sequence

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    limit of a sequence

    if a>0 and  x_{1}>0

    and  x_{n+1} = \frac{1}{2}(x_{n} +\frac{a}{x_{n}})

    find the limit of the sequence as n goes to infinity
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    Quote Originally Posted by xalk View Post
    if a>0 and  x_{1}>0

    and  x_{n+1} = \frac{1}{2}(x_{n} +\frac{a}{x_{n}})

    find the limit of the sequence as n goes to infinity
    i assume you have already proven that a (finite) limit exists?

    note that \lim_{n \to \infty}x_n = \lim_{n \to \infty}x_{n + 1}, call this limit L. Then we have


    \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} \frac 12 \left( x_n +\frac a{x_n} \right)

    \Rightarrow L = \frac 12 \left( L +\frac aL \right)

    Now find L
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    o.k thanks i have done that L= sqrt(a) or L= -sqrt(a)

    Now which is the limit of the function ,because up to now we assumed that the sequence converges we have not proved that the sequence converges
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    Quote Originally Posted by xalk View Post
    o.k thanks i have done that L= sqrt(a) or L= -sqrt(a)
    Now which is the limit of the function ,because up to now we assumed that the sequence converges we have not proved that the sequence converges
    The terms of the sequence are all positive.
    So which is it?

    To prove convergence, show that the sequence is monotone and bounded.
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    Thank you, but i am not so sure why we must take the positive sqrt of a as the limit of the function because all the terms of the sequence are positive .Is there a theorem or another problem ,saying that if all the terms of the sequence are +ve then the limit of the function must be +ve??

    To prove convergence i tried hard with no result so a complete proof will be very mush appreciated it
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    It is known that if a positive sequence converges its limit is non-negative.
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    can this be proved? if ,yes how??
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    If the limit of a sequence is negative then almost all of its terms must be negative.
    Say \left( {s_n } \right) \to L < 0 then using the definition let \varepsilon  = \frac{{ - L}}{2}.
    \left( {\forall n \geqslant N} \right)\left[ {\left| {s_n  - L} \right| <\varepsilon } \right]\, \Rightarrow \,s_n  < \frac{L}{2} < 0
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    Good .So a proof of convergence is left
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    Quote Originally Posted by xalk View Post
    So a proof of convergence is left
    Show it is bounded and eventually monotone.
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    HOW??
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    Quote Originally Posted by xalk View Post
    HOW??
    That is your job. It is so messy that I have no desire to even try it.
    I am glad I donít have to do it. Like I said, it is your problem.
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    By considering the difference :
     x_{n}-x_{n+1} = \frac{(x_{n}-\sqrt{a})(x_{n}+\sqrt{a})}{2x_{n}} we can study the monotony of the sequence .

    But in this difference all terms are positive except the term  x_{n}-\sqrt{a} which can be positive or -ve for all values of nεN.

    At the same time if we can prove whether the term  x_{n}-\sqrt{a} is +ve or -ve we will have proved whether the sequence is bounded above or bellow by sqrt(a).

    So how do we that ,if anyone knows please help.
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  14. #14
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    You can go here.
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    Thank you ,but i cannot see how you get:

    s_n^2 = x + (s_{n-1} - xs_{n-1})^2
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