# Evaluating $\lim\limits_{n\rightarrow \infty} \frac1{n^2}\ln \left( \frac{(n!)^n}{(0!1!2!…n!)^2} \right)$

Evaluating $$\lim\limits_{n\rightarrow \infty} \frac1{n^2}\ln \left( \frac{(n!)^n}{(0!1!2!…n!)^2} \right)$$

I’m not quite sure where to start in evaluating this. Some pointers, or a solution, would greatly be appreciated.

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No need to reference Barnes G-functions. These are integers after all.

$$\log{\left ( \frac{(n!)^n}{(0!1!2!…n!)^2} \right )} = n \log{n!} – 2 \sum_{k=0}^n \log{k!}$$

Now

$$\sum_{k=0}^n \log{k!} = \sum_{k=0}^n \log{\Gamma(k+1)} = \sum_{k=1}^{n+1} \log{\Gamma(k)}$$

Now use Euler-Maclurin:

\begin{align}\sum_{k=1}^{n+1} \log{\Gamma(k)}&= \int_0^{n+1} dx \, \log{\Gamma(x)} + \frac12 \left [\log{\Gamma(n+1)} – \log{\Gamma(1)} \right ] + o(n)\end{align}

Now, we use Raabe’s integral, which is

$$\int_{k}^{k+1} dx \, \log \Gamma(x) = \frac12 \log{(2 \pi)} + k \log{k}-k$$

so that

\begin{align} \int_{0}^{n+1} dx \, \log{\Gamma(x)} &= \sum_{k=0}^n \int_{k}^{k+1} dx \, \log{\Gamma(x)}\\ &= \frac{n+1}{2} \log (2 \pi) – \frac{n (n+1)}{2} + \sum_{k=1}^n k \log{k} \\ &= \frac{n+1}{2} \log (2 \pi) – \frac{n (n+1)}{2} + \log{(1^1 2^2 \cdots n^n)} \\ &= \frac{n+1}{2} \log (2 \pi) – \frac{n (n+1)}{2} + \log{\left [\frac{n!^n}{(n-1)!(n-2)!^2 \cdots 2^{n-2} 1^{n-1}} \right ]} \\ &= \frac{n+1}{2} \log (2 \pi) – \frac{n (n+1)}{2} + n \log{n!} – \sum_{k=1}^{n-1} \log{k!}\end{align}

Putting this all together, we find that

$$2 \sum_{k=1}^{n} \log{k!} = \frac{n+1}{2} \log (2 \pi) – \frac{n (n+1)}{2} + n \log{n!} + \frac{3}{2} \log{n!} +o(n)$$

so that, finally, using the fact that Stirling’s approximation is equivalent to $\log{n!} = n \log{n} – n + o(n)$,

$$\lim_{n \to \infty} \frac1{n^2} \left ( n \log{n!} – 2 \sum_{k=1}^{n} \log{k!} \right ) = \frac12$$

Raabe’s integral is actually not all that hard to evaluate. Rewrite as

$$\int_k^{k+1} dx \, \log{\Gamma(x)} = \int_0^1 dx \, \log{\Gamma(x+k)}$$

$$\log{\Gamma(x+k)} = \log{\Gamma(x)} + \sum_{m=0}^{k-1} \log{(x+m)}$$

Thus,

\begin{align}\int_k^{k+1} dx \, \log{\Gamma(x)} &= \int_0^1 dx \, \log{\Gamma(x)} + \sum_{m=0}^{k-1} \int_0^1 dx \, \log{(x+m)}\\ &= \int_0^1 dx \, \log{\Gamma(x)} + \sum_{m=0}^{k-1} [(m+1) \log{(m+1)} – m \log{m} – (m+1)+m] \\ &= \int_0^1 dx \, \log{\Gamma(x)} + k \log{k} – k \end{align}

To evaluate the integral on the RHS, use the duplication formula:

$$\Gamma \left ( \frac{x}{2} \right ) \Gamma \left ( \frac{x+1}{2} \right ) = 2^{1-x} \sqrt{\pi} \Gamma(x)$$

so that

$$\log{\Gamma(x)} = \log{\left [\frac{\Gamma \left ( \frac{x}{2} \right ) \Gamma \left ( \frac{x+1}{2} \right )}{\sqrt{\pi} 2^{1-x}} \right ]} = -\frac12 \log{\pi} – \log{2} + \log{\Gamma \left ( \frac{x}{2} \right )} + \log{\Gamma \left ( \frac{x+1}{2} \right )} + x \log{2}$$

Thus,

\begin{align}\int_0^1 dx \, \log{\Gamma(x)} &= -\frac12 \log{(2 \pi)} + \int_0^1 dx \, \log{\Gamma \left ( \frac{x}{2} \right )} + \int_0^1 dx \, \log{\Gamma \left ( \frac{x+1}{2} \right )}\\ &= -\frac12 \log{(2 \pi)} + 2 \int_0^{1/2} dx \, \log{\Gamma(x)} + 2 \int_{1/2}^1 dx \, \log{\Gamma(x)} \\ &= -\frac12 \log{(2 \pi)} + 2 \int_0^{1} dx \, \log{\Gamma(x)} \end{align}

The result follows.

Since
$$\int_1^n\log(x)\,\mathrm{d}x\le\log(n!)\le\int_1^{n+1}\log(x)\,\mathrm{d}x\tag{1}$$
we have that
$$\log(n!)=n\log(n)-n+O(\log(n))\tag{2}$$
therefore,
$$\log\left(\frac{n^{n-1}}{n!}\right)=n+O(\log(n))\tag{3}$$

Note that
$$\left.\frac{(n!)^n}{(0!1!2!\cdots n!)^2}\middle/\frac{((n-1)!)^{n-1}}{(0!1!2!\cdots(n-1)!)^2}\right. =\frac{n^{n-1}}{n!}\tag{4}$$
Applying $(3)$, we have
\begin{align} \log\left(\frac{(n!)^n}{(0!1!2!\cdots n!)^2}\right) &=\sum_{k=1}^n\log\left(\frac{k^{k-1}}{k!}\right)\\ &=\sum_{k=1}^n(k+O(\log(k)))\\ &=\frac{n^2}2+O(n\log(n))\tag{5} \end{align}
Therefore,
$$\lim_{n\to\infty}\frac1{n^2}\log\left(\frac{(n!)^n}{(0!1!2!\cdots n!)^2}\right)=\frac12\tag{6}$$

1). Stirling’s approximation and Stirling-like asymptotic series for Barnes G- function here at $(14)$ http://mathworld.wolfram.com/BarnesG-Function.html

Therefore,

$$\lim_{n\rightarrow \infty} \dfrac{\displaystyle \ln \left ( \frac{(n!)^n}{(0!1!2!…n!)^2} \right )}{n^2}=\frac{1}{2}.$$

2). Apply Stolz–Cesàro theorem for an elementary solution.

3). Combine the squeeze theorem, Stirling’s approximation and the Riemann sums.

4). Make use of the Euler–Maclaurin formula.