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Math Help - convergence

  1. #1
    mms
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    convergence

    <br />
\mathop {\lim }\limits_{n \to \infty } \sqrt[{n^2 }]{{\prod\limits_{i = 1}^n {\left( \begin{gathered}<br />
n \hfill \\<br />
i \hfill \\ <br />
\end{gathered} \right)} }}<br /> <br />
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  2. #2
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    Quote Originally Posted by mms View Post
    <br />
\mathop {\lim }\limits_{n \to \infty } \sqrt[{n^2 }]{{\prod\limits_{i = 1}^n {\left( \begin{gathered}<br />
n \hfill \\<br />
i \hfill \\ <br />
\end{gathered} \right)} }}<br /> <br />
    I get the answer to be \sqrt e\approx 1.6487.

    Start by checking that \int_0^1\!\!\!x\ln x\,dx = -1/4 (easy integration by parts). Approximating this integral by a Riemann sum, you see that \frac1n\sum_{k=1}^n\frac kn\ln\Bigl(\frac kn\Bigr)\sim -\frac14 (for large n). Therefore \frac1{n^2}\sum_{k=1}^nk\ln k\sim -\frac14 + \frac{(n+1)\ln n}{2n} .....(*).

    Next,
    \begin{aligned}\prod_{k=1}^n{n\choose k} &= \frac n1\cdot\frac{n(n-1)}{1\cdot2}\cdot \frac{n(n-1)(n-2)}{1\cdot2\cdot3}\cdots \\ &= n^{n-1}(n-1)^{n-3}(n-2)^{n-5}\cdots 2^{-n+3} = \prod_{k=1}^nk^{2k-n-1}.\end{aligned}

    Take logs to see that
    \begin{aligned}\ln\left(\sqrt[n^2 ]{\prod_{k = 1}^n {n\choose k}}\right) &= \sum_{k=1}^n\frac{2k-n-1}{n^2}\ln k \\ &\sim -\frac12 + \frac{n+1}n\ln n - \frac{n+1}{n^2}\sum_{k=1}^n\ln k\qquad\text{(from (*))} \\ &= -\frac12 + \frac{n+1}n\ln n - \frac{n+1}{n^2}\ln(n!) \\ &\sim -\frac12 + \frac{n+1}n\ln n - \frac{n+1}n(\ln n-1)\quad\ \text{(from Stirling's formula)}.\end{aligned}
    This converges to 1/2 as n→∞, which leads to the result stated at the beginning.
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