Asymptotic Methods

Oct 2007
722
168
Show that

\(\displaystyle 1^1 2^2 3^3 \cdots n^n \sim Cn^{\tfrac{1}{2}n^2 + \tfrac{1}{2}n + \tfrac{1}{12}}e^{-\tfrac{1}{4}n^2} \qquad \textrm{ as } n \to \infty\)



Moderator edit: Approved Challenge question.
 
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Drexel28

MHF Hall of Honor
Nov 2009
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Berkeley, California
Show that

\(\displaystyle 1^1 2^2 3^3 \cdots n^n \sim Cn^{\tfrac{1}{2}n^2 + \tfrac{1}{2}n + \tfrac{1}{12}}e^{-\tfrac{1}{4}n^2} \qquad \textrm{ as } n \to \infty\)
I don't want to give the solution because I have seen this before. You do realize this is "commonly" known? The value of \(\displaystyle C\) is usually denoted \(\displaystyle A\) and is known as the Glaisher-Kinkelin constant.
 

Laurent

MHF Hall of Honor
Aug 2008
1,174
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Paris, France
It wouldn't be a very cool proof, but you can let \(\displaystyle u_n=\log(1^12^2\cdots n^n)-\frac 12 n^2\log n-\frac 12 n\log n+\frac 14 n^2\) and check that \(\displaystyle u_n-u_{n-1}=\frac{1}{12n}+O\left(\frac{1}{n^2}\right)\) (very simple but lengthy, or Maple can do it for you). Since \(\displaystyle \sum_{k=2}^n\frac{1}{k} = \log n -\gamma+o(1)\) and \(\displaystyle \sum_k\frac{1}{k^2}\) converges, we deduce \(\displaystyle u_n=\sum_{k=2}^n (u_k-u_{k-1})=\frac{1}{12}\log n-C+o(1)\). Taking exponentials gives the result...