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How to prove if the following limit exists?

If it exists, what's the value?

$$\lim\limits_{n\to\infty}\dfrac{\sum\limits_{k=0}^n\log\binom{n}{k}}{n^2}$$

Thanks!

Björn Friedrich
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2 Answers2

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Notice that we can write

$$ \frac{1}{n^2} \sum_{k=1}^{n} \log \binom{n}{k} = \frac{1}{n}\sum_{k=1}^{n} \left( \frac{2k-1}{n} - 1 \right) \log\left(\frac{k}{n}\right). $$

Taking $n \to \infty$, this converges to the integral

$$ \int_{0}^{1} (2x-1)\log x \, \mathrm{d}x = \frac{1}{2}. $$

Sangchul Lee
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    [+1] Excellent! I just obtained the same value using Stirling approximation. Confirmed also by numerical simulation. – Jean Marie May 04 '17 at 13:42
  • @JeanMarie, Thank you! – Sangchul Lee May 04 '17 at 13:47
  • Could you please elaborate on why $\dfrac{1}{n}\log{\binom{k}{n}} = \left(\dfrac{2k-1}{n}-1\right)\log\left(\dfrac{k}{n}\right)$? I wonder why do you substitute binomial coefficient with fraction. – Ramil May 04 '17 at 13:48
  • @Ramil, Your computation is pretty close to what I have done there. Notice that the sum above equals $$ \frac{1}{n^2} \sum_{k=1}^{n} (2k - 1 - n) \log k = \frac{1}{n^2} \sum_{k=1}^{n} (2k - 1 - n) \log \left(\frac{k}{n}\right) + \frac{\log n}{n^2} \sum_{k=1}^{n} (2k - 1 - n). $$ Computation shows that the latter sum vanishes, hence yields the identity. – Sangchul Lee May 04 '17 at 13:51
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    For improper integrals, the convergence of Riemann sums is not automatic, but it seems to be OK here. See https://math.stackexchange.com/questions/1744250/convergence-of-riemann-sums-for-improper-integrals for example. – Barry Cipra May 04 '17 at 13:56
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    @SangchulLee Thanks! Your solution is correct indeed, I found a mistake in my computations. – Ramil May 04 '17 at 13:59
  • @Barry Cipra, I agree that it required elaboration. If we are forced to use only preliminary level of analysis, justification step may be pesky. Otherwise, we can simply appeal to the dominated convergence theorem together with the estimate $$ \left| \sum_{k=1}^n \left( \frac{2k-1}{n} - 1 \right) \log\left(\frac{k}{n}\right) \mathbf{1}_{(\frac{k-1}{n}, \frac{k}{n}]}(x) \right| \leq \left| \log x \right| $$ valid for $x \in (0, 1]$. – Sangchul Lee May 04 '17 at 14:00
  • Thank you so much! I've learned a lot. –  May 04 '17 at 14:22
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}{\sum_{k = 0}^{n}\ln\pars{n \choose k} \over n^{2}} & = \lim_{n \to \infty} {\sum_{k = 1}^{n + 1}\ln\pars{n + 1 \choose k} - \sum_{k = 1}^{n}\ln\pars{n \choose k} \over \pars{n + 1}^{2} - n^{2}} = \lim_{n \to \infty} {\sum_{k = 1}^{n}\ln\pars{{n + 1 \choose k}/{n \choose k}} \over 2n + 1} \\[5mm] & = {1 \over 2}\,\lim_{n \to \infty} {\sum_{k = 1}^{n}\bracks{\ln\pars{n + 1} - \ln\pars{n - k + 1}} \over n + 1/2} \\[5mm] & = {1 \over 2}\,\lim_{n \to \infty}{n\ln\pars{n + 1} - \ln\pars{n!} \over n + 1/2} \\[5mm] & = {1 \over 2}\,\lim_{n \to \infty}\bracks{\vphantom{\Large A}% \pars{n + 1}\ln\pars{n + 2} - n\ln\pars{n + 1} - \ln\pars{n + 1}} \\[5mm] & = {1 \over 2} \lim_{n \to \infty}\bracks{\pars{n + 1}\ln\pars{1 + 2/n \over 1 + 1/n}} = \bbx{1 \over 2} \end{align}

Felix Marin
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    I realized I could use Stolz theorem twice and solved it easily several hours after I posted the question... Thank you all the same. –  May 06 '17 at 09:42
  • @Maristie Thanks. When I saw the $@\texttt{Sangchul Lee}$ answer, I thought Stoltz-Ces$\mathrm{\grave{a}}$ro (SC) doesn't work because $@\texttt{Sangchul Lee}$ didn't use it. However, after awhile I realized that SC indeed works very fine. – Felix Marin May 06 '17 at 20:00
  • I admit that I am weak at utilizing Cesaro-Stolz theorem. Nice solution! (+1) – Sangchul Lee May 07 '17 at 17:37
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    @SangchulLee Thanks. Indeed, I believe you are quite strong. – Felix Marin May 07 '17 at 22:16