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 $o(n)$?