Help calculate this limit,about double factorial. $$
\lim_ {n\to\infty} \dfrac {\left [\left (2n-1\right)!! \right] ^ {1/ {2n}}} {\left [\displaystyle\prod_ {k=1} ^ {n} (2k-1)!! \right] ^ {1/ {n^2}}}$$

The numerator is $\int_0^1 x\log(x)dx=-1/4$
so $\lim_{n\to\infty}\frac{1}{n}(\sum_{k=1}^n \frac{2k-1}{2n-1}\log(\frac{2k-1}{2n-1}))=-\frac14$
so $\lim_{n\to\infty}(\prod_{k=1}^n(\frac{2k-1}{2n-1})^{2k-1})^{\frac{1}{n(2n-1)}}=e^{-\frac14}$
