Question

If limit $\underset{n\to \infty }{lim}{n}^{\frac{-1}{2}(1+\frac{1}{n})}(1^1{2}^2.3^3....n^n{)}^{\frac{1}{n^2}}=L$,
then $-\ln L$ is

Answer 1

$\ln L=\underset{n\to \infty }{lim}-\frac{1}{2}(1+\frac{1}{n})\ln n+\frac{1}{n^2}{\sum }_{k=1}^{n}k\ln k$
$\ln L={\int }_0^1\ln xdx=\frac{-1}{4}$
$L=e^{\frac{-1}{4}}$