Limit $\lim _{n \to \infty} \frac {\sqrt[2n] {(2n-1)!!}} {\sqrt [n^2] a_n}$

Question

Let us consider the sequence $(a_n)_{n \ge 1}$ defined as follows: $a_1 = 1$ and $a_{n+1}=(2n+1)!! a_n, \forall n \ge 1$. I had to compute the limit: $$\lim _{n \to \infty} \frac {\sqrt[2n] {(2n-1)!!}} {\sqrt [n^2] a_n}$$ that Jack D'Aurizio's solution below shows to be $e^{1/4}$ through Riemann sums.
I am wondering about different/simpler approaches: is it possible to tackle such limit through the Stoltz-Cesàro theorem, for instance?

Answer 1

^{This is the original approach through Riemann sums, with an extended explanation here.}

$$ a_n = \prod_{k=1}^{n}(2k-1)^{n+1-k} \tag{1}$$ leads to $$ \frac{1}{n^2}\log a_n = \frac{1}{n^2}\sum_{k=1}^{n}(n+1-k)\log(2k-1) \tag{2}$$ and to: $$ \frac{1}{2n}\log(2n-1)!! = \frac{1}{2n}\sum_{k=1}^{n}\log(2k-1) \tag{3} $$ hence $(2)-(3)$ equals $$ \frac{1}{n}\sum_{k=0}^{n-1}\left[\frac{k+1}{n}-\frac{1}{2}\right]\log(2n-2k-1) \tag{4}$$ that for large values of $n$ converges to $$ \int_{0}^{1}\left(x-\frac{1}{2}\right)\log(2-2x)\,dx=-\frac{1}{4}.\tag{5} $$ The given limit is so $\color{red}{\large e^{1/4}}.$