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Math Help - Limit of infinite product

  1. #1
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    Limit of infinite product

    Hello. I need help in calculating the limit.
    Lim((sqrt((n+1)*(n+2)*(n+3)*...*(2n-1)*2*n) of n)/n), n=infinity.
    Surely, it exists.
    I can't understand how I shall express the root of the product in the nominator or should I. I can understand that there must be k*n in the nominator, but I don't know how to find this k. simple product of n is not suitable.
    Great thanks for everybody who will try to help me.
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  2. #2
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    Quote Originally Posted by Serg777 View Post
    Hello. I need help in calculating the limit.
    Lim((sqrt((n+1)*(n+2)*(n+3)*...*(2n-1)*2*n) of n)/n), n=infinity.
    Surely, it exists.
    I can't understand how I shall express the root of the product in the nominator or should I. I can understand that there must be k*n in the nominator, but I don't know how to find this k. simple product of n is not suitable.
    Great thanks for everybody who will try to help me.
    Just for clarifaction, are you trying to find

    \lim_{n \to \infty} \frac{\left((n+1)(n+2) \cdots (2n-1)2n\right)^{\frac{1}{n}}}{n}
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  3. #3
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    Yes.I am.
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  4. #4
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    Try this

    <br />
\lim_{n \to \infty} \frac{(n+1)^{\frac{1}{n}} (n+2)^{\frac{1}{n}} \cdots (2n-1)^{\frac{1}{n}} (2n)^{\frac{1}{n}} }{n^{\frac{1}{n}} \qquad \;\;n^{\frac{1}{n}} \cdots \qquad \;\;n^{\frac{1}{n}} \qquad \;\;n^{\frac{1}{n}} }

    then group separately. So you're really looking at

    \lim_{n \to \infty} \left( 1 + \frac{i}{n}\right)^{\frac{1}{n}}\;\;\; i = 1 \cdots n
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  5. #5
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    Thank you VERY MUCH
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  6. #6
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    The last limit is the limit of the sum. Correct? Can you explain in full how to group?
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  7. #7
    Moo
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    Quote Originally Posted by Serg777 View Post
    The last limit is the limit of the sum. Correct? Can you explain in full how to group?
    \lim_{n \to \infty} \frac{(n+1)^{\frac{1}{n}} (n+2)^{\frac{1}{n}} \cdots (2n-1)^{\frac{1}{n}} (2n)^{\frac{1}{n}} }{n^{\frac{1}{n}} \qquad \;\;n^{\frac{1}{n}} \cdots \qquad \;\;n^{\frac{1}{n}} \qquad \;\;n^{\frac{1}{n}} }<br />
    =\lim_{n \to \infty} \left(\frac{n+1}{n}\right)^{1/n} \cdots \left(\frac{2n-1}{n}\right)^{1/n} \left(\frac{2n}{n}\right)^{1/n}



    In fact, it's the limit of \lim_{n \to \infty} {\color{red}\prod_{i=1}^n} \left(1+\frac in\right)^{1/n}, the limit of the product
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  8. #8
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    I thought it was smth more complicated about grouping. Is it possible to reduce to sum? Can we omit everything except numbers and 1/(n+1)? By grouping (n+1)*(1+2/(1+n))^(1/n).....
    Doing this we get (1+1/(n+1))*(1+2/(n+1).....). Of course, root of n.

    What do you think?
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  9. #9
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    Because I have no idea how to find the limit of product.
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  10. #10
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    Following up on Moo's idea, we have that the product equals \exp \left\{ \frac{1}{n}\sum\limits_{i=1}^{n}{\ln \left( 1+\frac{i}{n} \right)} \right\}\to \exp \left( \int_{1}^{2}{\ln x\,dx} \right) as n\to\infty.

    There're a few ways of solving this last integral, the standar one, it's to apply integration by parts.
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  11. #11
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    Thanks, fellows. You are great.
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