1. ## Asymptotic Methods

Show that

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

Moderator edit: Approved Challenge question.

Show that

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

3. Originally Posted by Drexel28
You do realize this is "commonly" known?
Yeah totally that's why I posted it in a maths challenge section.

4. Spoiler:

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