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Math Help - Limits / Series question

  1. #1
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    Limits / Series question

    hi

    Let
    P = [(2^3-1)/(2^3+1)][(3^3-1)/(3^3+1)]........[(n^3-1)/(n^3+1)]
    n=2,3,4,......

    Find Limit P when N tends to infinity.
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  2. #2
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    Quote Originally Posted by champrock View Post
    hi

    Let
    P = [(2^3-1)/(2^3+1)][(3^3-1)/(3^3+1)]........[(n^3-1)/(n^3+1)]
    n=2,3,4,......

    Find Limit P when N tends to infinity.
    Factorise \frac{n^3-1}{n^3+1} as \frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}, and notice that n^2+n+1 = (n+1)^2 - (n+1) + 1. You will then find that you have a telescoping product.
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  3. #3
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    what should i do with n^2 - n + 1 ?

    I was able to cancel the (n-1)/(n+1) terms. but dont know what to do with n^2 - n + 1 and n^2 + n + 1
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  4. #4
    MHF Contributor chisigma's Avatar
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    You easily undestand if you write the sequence of terms...

    P= \prod_{n=2}^{\infty} \frac{(n-1)\cdot (n^{2} + n + 1)}{(n+1)\cdot (n^{2} - n + 1)}= \frac {1\cdot 7}{3\cdot 3}\cdot \frac {2\cdot 13}{4\cdot 7}\cdot \frac {3\cdot 21}{5\cdot 13}\cdot \frac {4\cdot 31}{6\cdot 21}\cdot  \dots

    ... and it is evident you can simplify both in numerator and denominator [3 with 3, 4 with 4, ... , 7 with 7, 13 with 13, 21 with 21,...]

    Finally is...

    P=\frac{2}{3}

    Very nice!...

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by champrock View Post
    what should i do with n^2 - n + 1 ?

    I was able to cancel the (n-1)/(n+1) terms. but dont know what to do with n^2 - n + 1 and n^2 + n + 1
    Use the hint I gave before: n^2+n+1 = (n+1)^2 - (n+1) + 1. That tells you that the term n^2+n+1 in the numerator of each fraction cancels with the term (n+1)^2 - (n+1) + 1 in the denominator of the next one.
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  6. #6
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    I think Only 1/3 is left. All the other terms cancel out. How are you getting 2/3
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  7. #7
    MHF Contributor chisigma's Avatar
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    The term of order n is...

    a_{n}= \frac{(n-1)\cdot (n^{2} +n + 1)}{(n+1)\cdot (n^{2} - n + 1)}

    For n=3 is...

    a_{3}= \frac{2\cdot 13 }{4\cdot 7}

    Kind regards

    \chi \sigma
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