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Math Help - Limit when n-> infinity

  1. #1
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    Limit when n-> infinity

    \lim_{n->\infty}\frac{qn+1}{qn}\frac{qn+p+1}{qn+p}\frac{qn  +2p+1}{qn+2p}...\frac{qn+np+1}{qn+np}

    p\in N-\begin{Bmatrix}0,<br />
 & 1<br />
\end{Bmatrix}<br />
,q\in N^\ast
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  2. #2
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    Quote Originally Posted by m3th0dman View Post
    \lim_{n->\infty}\frac{qn+1}{qn}\frac{qn+p+1}{qn+p}\frac{qn  +2p+1}{qn+2p}...\frac{qn+np+1}{qn+np}

    p\in N-\begin{Bmatrix}0,<br />
 & 1<br />
\end{Bmatrix}<br />
,q\in N^\ast
    I tend to be mistrustful of "big O" arguments, but I think this one is okay.

    Notice that \frac{qn+kp+1}{qn+kp} = 1 + \frac1{qn+kp}, and \ln\Bigl(1 + \frac1{qn+kp}\Bigr) = \frac1{qn+kp} + O(n^{-2}). Therefore

    \begin{aligned}\sum_{k=0}^n\ln\Bigl(1 + \frac1{qn+kp}\Bigr) &= \sum_{k=0}^n\frac1{qn+kp} + O(n^{-1})\\ &= \frac1n\sum_{k=0}^n\frac1{q+\frac knp} + O(n^{-1})\\ &\to \int_0^1\frac1{q+xp}\,dx\qquad\text{(limit of Riemann sum)}\\ &= \tfrac1p\ln\bigl(1+\tfrac pq\bigr).\end{aligned}

    Taking exponentials, \lim_{n\to\infty}\prod_{k=0}^n \frac{qn+kp+1}{qn+kp} = \bigl(1+\tfrac pq\bigr)^{1/p}.
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    Quote Originally Posted by Opalg View Post
    I tend to be mistrustful of "big O" arguments, but I think this one is okay.

    Notice that \frac{qn+kp+1}{qn+kp} = 1 + \frac1{qn+kp}, and \ln\Bigl(1 + \frac1{qn+kp}\Bigr) = \frac1{qn+kp} + O(n^{-2}). Therefore

    \begin{aligned}\sum_{k=0}^n\ln\Bigl(1 + \frac1{qn+kp}\Bigr) &= \sum_{k=0}^n\frac1{qn+kp} + O(n^{-1})\\ &= \frac1n\sum_{k=0}^n\frac1{q+\frac knp} + O(n^{-1})\\ &\to \int_0^1\frac1{q+xp}\,dx\qquad\text{(limit of Riemann sum)}\\ &= \tfrac1p\ln\bigl(1+\tfrac pq\bigr).\end{aligned}

    Taking exponentials, \lim_{n\to\infty}\prod_{k=0}^n \frac{qn+kp+1}{qn+kp} = \bigl(1+\tfrac pq\bigr)^{1/p}.
    Thank you very much. This is the correct answer (written at the end of the Book). Unfortunately I am not familiar with Big O notation, so I will have to chew on it a little.
    I would be grateful if you could explain it without "Big O" notation, or if easier you could explain the "Big O" notation.
    I understand that is a function that is not important in finding the answer, written to simplfy things.
    In this is case it depends on 1/n^2, so I supouse it goes to 0 when n-> infinity.
    Last edited by m3th0dman; April 15th 2009 at 01:02 PM.
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  4. #4
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    Quote Originally Posted by m3th0dman View Post
    I would be grateful if you could explain it without "Big O" notation, or if easier you could explain the "Big O" notation.
    I understand that is a function that is not important in finding the answer, written to simplify things.
    In this is case it depends on 1/n^2, so I suppose it goes to 0 when n-> infinity.
    The big O notation is just a succinct way of estimating orders of magnitude. For this problem, you can make the estimate explicit by using Taylor's theorem with remainder for the function ln(1+x). This tells you that, for 0<x<1, the difference between ln(1+x) and x is less than \tfrac12x^2. Put x= \frac1{qn+kp} to see that the difference between \ln\Bigl(1 + \frac1{qn+kp}\Bigr) and \frac1{qn+kp} is less than \frac1{2q^2n^2}. Adding n+1 such terms together, you find that the difference between \sum_{k=0}^n\ln\Bigl(1 + \frac1{qn+kp}\Bigr) and \sum_{k=0}^n\frac1{qn+kp} is less than \frac{n+1}{2q^2n^2}. This difference goes to 0 as n goes to infinity.
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  5. #5
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    Quote Originally Posted by Opalg View Post
    The big O notation is just a succinct way of estimating orders of magnitude. For this problem, you can make the estimate explicit by using Taylor's theorem with remainder for the function ln(1+x). This tells you that, for 0<x<1, the difference between ln(1+x) and x is less than \tfrac12x^2. Put x= \frac1{qn+kp} to see that the difference between \ln\Bigl(1 + \frac1{qn+kp}\Bigr) and \frac1{qn+kp} is less than \frac1{2q^2n^2}. Adding n+1 such terms together, you find that the difference between \sum_{k=0}^n\ln\Bigl(1 + \frac1{qn+kp}\Bigr) and \sum_{k=0}^n\frac1{qn+kp} is less than \frac{n+1}{2q^2n^2}. This difference goes to 0 as n goes to infinity.
    I suppose that Taylor's formula is applied this way:
    <br />
ln(1+x)=ln(a+1)+\frac{x-a}{1+a}-\frac{(x-a)^2}{2!(1+a)^2}+\frac{(x-a)^3}{3!(1+a)^4}-...<br />
    And we apply it for a=0:
    <br />
ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3!}-\frac{x^4}{4!}+...<br />
    and
    <br />
\frac{x^2}{2} \geq -\frac{x^2}{2}+\frac{x^3}{3!}-\frac{x^4}{4!}+...<br />
    for x between [0,1].

    I haven't studied Taylor's Theorem (or formula) at school, so I don't know if I had applied correctly.

    Assuming what have I written is correct, I understood everything. Again I thank you very much.
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