Asymptotics of product of harmonic numbers Consider the product 
$$p_n = \prod_{k=1}^n H_k$$
of $n$ successive harmonic numbers $H_k=\sum_{i=1}^k 1/i$.
The sequence of the $p_n$ is listed in OEIS as A097423/A097424.
I am looking for the asymptotic behaviour of $p_n$ as $n\to\infty$.
My first attemps are based on the asymptotic behaviour of the harmonic number itself
$$H_n \simeq \log(n) + \gamma +\frac{1}{2n}-\frac{1}{12 n^2}+...$$
Defining     
$$q_{a,1}(n) = \prod_{k=1}^n \frac{H_k}{\log(k) + \gamma +\frac{1}{2k}}$$
$$q_{a,2}(n) = \prod_{k=1}^n \frac{H_k}{\log(k) + \gamma +\frac{1}{2k}-\frac{1}{12 k^2}}$$
I found numerically that 
$$q_{a,1}(10^3)\simeq 0.903394162407764$$
$$q_{a,2}(10^3)\simeq 1.006552015286574$$
Questions   
1) Are there closed expressions for the $q_{a,i}(\infty)$ in terms of known constants?
2) What can be said about the asymptotic behaviour of the products
$$r_{a,1}(n) = \prod_{k=1}^n \left( \log(k) + \gamma +\frac{1}{2k}\right)$$
$$r_{a,2}(n) = \prod_{k=1}^n \left(\log(k) + \gamma +\frac{1}{2k}-\frac{1}{12 k^2}\right)$$
 A: The obvious thing is to
extract the $\ln(n)$
from the product.
$\begin{array}\\
p_n 
&= \prod_{k=1}^n H_k\\
&= \prod_{k=1}^n \left( \log(k) + \gamma +f(1/k)\right)
\qquad\text{where }f(x) = x/2+O(x^2)\\
&= \prod_{k=2}^n  \log(k)\prod_{k=1}^n \left( 1 + (\gamma +f(1/k))/\log(k)\right)
\qquad\text{start the first product at 2 otherwise it is zero}\\
&=g(n)h(n)\\
\end{array}
$
Note:
This estimate changed.
$\begin{array}\\
g(n)
&=\prod_{k=2}^n  \log(k)\\
\text{so}\\
\log(g(n))
&=\sum_{k=2}^n  \log\log(k)\\
&\approx\int_{k=2}^n  \log\log(x)dx\\
&=  x\log\log(x)|_{2}^n-\int_2^n dx/\ln(x)\\
&\approx n\log\log(n)-n/\log(n)\\
\end{array}
$
and
$h(n)
=\prod_{k=1}^n \left( 1 + (\gamma +f(1/k))/\log(k)\right)
$
so
$\begin{array}\\
\ln(h(n))
&=\sum_{k=1}^n \log\left( 1 + (\gamma +f(1/k))/\log(k)\right)\\
&\approx\sum_{k=1}^n (\gamma +f(1/k))/\log(k))\\
&\approx\dfrac{n\gamma}{\log(n)}+O(\log\log(n))
\\
\end{array}
$
so,
finally,
$\log(p_n)
\approx n\log\log(n)-n/\log(n)+\dfrac{n\gamma}{\log(n)}+O(\log\log(n))
= n\log\log(n)-(1-\gamma)n/\log(n)+O(\log\log(n))
$.
This could be tested.
$\log(g(n))$
could have more terms from
a better estimate for
$\sum_{k=1}^n  \log\log(k)$.
