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A limit!

Evaluate the limit:

    \[\ell = \lim_{m\rightarrow +\infty}\sqrt[m]{m+1}\cdot\sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}}\]

Solution

Let a_m=\sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}}. Then,

    \[\log a_m = \log \sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}} = \frac{1}{m^2+m} \sum_{k=1}^{m} \log \binom{m}{k}\]

It follows by Stolz–Cesàro that

    \begin{align*} \lim_{m \rightarrow +\infty} \log a_m &= \lim_{m \rightarrow +\infty} \frac{1}{m^2+m} \sum_{k=1}^{m} \log \binom{m}{k} \\ &=\lim_{m \rightarrow +\infty} \frac{\sum \limits_{k=1}^{m+1} \log \binom{m+1}{k} - \sum \limits_{k=1}^{m} \log \binom{m}{k}}{(m+1)^2 + (m+1) -(m^2+m)} \\ &=\lim_{m \rightarrow +\infty} \frac{1}{2m+2} \sum_{k=1}^{m} \log \frac{m+1}{m+1-k} \\ &= \lim_{m \rightarrow +\infty} \frac{\log \frac{(m+1)^m}{m!}}{2m+2}\\ &=\lim_{m \rightarrow +\infty} \frac{1}{2} \cdot \frac{m}{m+1} \cdot \log \frac{m+1}{\sqrt[m]{m!}} \\ &= \frac{1}{2} \cdot 1 \cdot \log e \\ &= \frac{1}{2} \end{align*}

In addition,

    \begin{align*} \lim_{m \rightarrow +\infty} \sqrt[m]{m+1} &= \lim_{m \rightarrow +\infty} \left ( m+1 \right )^{1/m} \\ &=\lim_{m \rightarrow +\infty} e^{\frac{\log m}{m+1}} \\ &=1 \end{align*}

Hence \ell = \sqrt{e}.

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1 Comment

  1. We can also invoke another technique; that of the double summation.

    First of all , we note that

        \[\sum_{k=1}^{n} \log k! = \sum_{k=1}^{n} \sum_{j=1}^{k} \log j = \sum_{j=1}^{n} (n+1-j) \log j\]

    Hence,

        \begin{align*} \sum_{k=1}^{m} \log \binom{m}{k} &= (m+1) \log m! - 2 \sum_{k=1}^{m} \log k! \\ &= \sum_{j=1}^{m} (2j-m-1) \log j \\ &= \sum_{j=1}^{m} (2j-m-1) \log \left(\frac{j}{m}\right) \end{align*}

    The last line follows from the fact that \sum \limits_{j=1}^{m} (2j-m-1) \log m = 0. Thus,

        \begin{align*} \frac{1}{m(m+1)}\sum_{k=1}^{m} \log \binom{m}{k} &= \frac{m}{m+1} \sum_{j=1}^{m} \left(\frac{2j-1}{m}-1\right) \log \left(\frac{j}{m}\right) \, \frac{1}{m} \\ &\longrightarrow \int_{0}^{1} (2x-1)\log x \, \mathrm{d}x = \frac{1}{2} \end{align*}

    However, justifying this convergence requires more than the usual Riemann sum argument. 🙂

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