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Math Help - Finding a limit of two given series

  1. #1
    Junior Member hoeltgman's Avatar
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    Finding a limit of two given series

    I got two series defined as following:

     u(n+1) = \dfrac{u(n) + v(n)}{2} \text{with u(0)=a}
     \dfrac{1}{v(n+1)} = \dfrac{1}{2}(\dfrac{1}{u(n)} + \dfrac{1}{v(n)}) \text{with v(0)=b}

    a and b are positive real numbers.

    These two series converge to the same limit. Which I'm supposed to find. I managed to prove that these two series are convergent and even adjacent. But I don't see how to get the limit. All I found out by inserting numbers was that the limit probably is  \sqrt{ab}
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  2. #2
    hpe
    hpe is offline
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    Quote Originally Posted by hoeltgman
    I got two series defined as following:

     u(n+1) = \dfrac{u(n) + v(n)}{2} \text{with u(0)=a}
     \dfrac{1}{v(n+1)} = \dfrac{1}{2}(\dfrac{1}{u(n)} + \dfrac{1}{v(n)}) \text{with v(0)=b}

    a and b are positive real numbers.

    These two series converge to the same limit. Which I'm supposed to find. I managed to prove that these two series are convergent and even adjacent. But I don't see how to get the limit. All I found out by inserting numbers was that the limit probably is  \sqrt{ab}
    You must assume that ab > 0. I guess you have already proved that \lim_{n \to \infty}u(n) = \lim_{n \to \infty}v(n). Call the limit U.
    Now from the definition,
    v(n+1) = \frac{2 u(n)v(n)}{u(n)+v(n)} = \frac{u(n)v(n)}{u(n+1)}.
    Therefore for all n,
    u(n+1)v(n+1) = u(n)v(n).
    By induction,
    u(n)v(n) = u(0)v(0) = ab
    for all n. Now take the limit:
    U^2 = ab and therefore U = \sqrt{ab}.
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  3. #3
    Junior Member hoeltgman's Avatar
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    Thx. I had tried finding some results through some sort of induction, but I always ended up with expressions that either far too complicated, or ten kilometers long.
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