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I would like to find an equivalent of:

$ u_{n}=\prod_{k=1}^{n} k^k $

I managed to find and asymptotic expansion of $ \ln(u_{n}) $ whose precision is $ o(n) $:

$ \ln(u_{n})=\frac{1}{2}n^2\ln(n)-\frac{n^2}{4}+\frac{1}{2}n\ln(n)+o(n)$

Indeed: $ \ln(u_{n}) \sim \frac{1}{2}n^2\ln(n)$

Introducing: $ v_{n}=\ln(u_{n})-\frac{1}{2}n^2\ln(n)$, we have:

$ v_{n}-v_{n-1} \sim -\frac{n}{2}$

So: $ v_{n} \sim \sum_{k=2}^nv_{k}-v_{k-1} \sim -\frac{n^2}{4}$

Introducing $ w_{n}=v_{n}+\frac{n^2}{4}$ I get:

... $ w_{n} \sim \sum_{k=2}^nw_{k}-w_{k-1} \sim\frac{1}{2}n\ln(n)$

And $x_{n}=w_{n}-\frac{1}{2}n\ln(n) \Longrightarrow x_{n}-x_{n-1}=o(1)\Longrightarrow x_{n}=o(n)$

As $ x_{n}=0+o(n) $ the method used for $v_{n}$ and $w_{n}$ becomes useless, so how can I determine that $o(n)$?

  • 2
    $u_n$ is known as [hyperfactorial](http://en.wikipedia.org/wiki/Factorial#Hyperfactorial) (see also, at [MathWorld](http://mathworld.wolfram.com/Hyperfactorial.html))2012-04-11

1 Answers 1

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If we wanted to we could rewrite it as

$\begin{array}{c l} 1^1\cdot 2^2\cdot 3^3\cdots n^n & =(1\cdot2\cdots n)\times (2\cdot3\cdots n)\times\cdots \big((n-1)\cdot n\big) \times n \\ & = \frac{n!}{0!}\times \frac{n!}{1!}\times \cdots \times \frac{n!}{(n-1)!} \\ & = \frac{n!^{\,n}}{G(n+1)} \end{array}$

where $G(\cdot)$ is the Barnes G function.

Anyway, with the Euler-Maclaurin formula we have

$\begin{array}{c l} \log u_n & = \sum_{k=1}^n k\log k \\ & = \color{Blue}{\int_1^n x\log xdx}+\color{DarkBlue}{\frac{1\log 1+n\log n}{2}}+\color{Purple}{\frac{-1/2}{2!}\big( (1+\log n)-(1+\log 1)\big)}+\color{Maroon}\cdots \\ & = \color{Blue}{\frac{n^2}{2}\left(\log n-\frac{1}{2}\right)+\frac{1}{4}}+\color{DarkBlue}{\frac{n\log n}{2}}-\color{Purple}{\frac{\log n}{4}}+\color{Maroon}\cdots \end{array}$