How can I be certain that this infinite product has no associated value?

Question

I have a product

$\prod_{n=2}^{\infty}{\ln{n}}$

which fails the standard test for convergence with

$\sum_{n=2}^{\infty}{\ln(\ln(x))}$

not converging. However, I am aware that that other seemingly divergent sums have associated values, such as the sum of all natural numbers, which comes from the Riemann zeta function. So how can I be sure that this indeed diverges?

As a side note, I noticed that the partial sums

$\lim_{k\rightarrow\infty}\sum_{n=2}^{k}{\ln(\ln(x))}$

appear to be converging on a line with the slope probably between $1<m<2$. I haven't been able to check this for very very large values, but it seems to hold true for the partial sums atleast $k < 2000$

Edit:

It recently came to my attention that I may be able assign a value with the zeta function by using the sum above, but I'm not sure how.

Answer 1

However, I am aware that that other seemingly divergent sums have associated values, such as the sum of all natural numbers, which comes from the Riemann zeta function. So how can I be sure that this indeed diverges?

Seeing how you understand that the sum diverges (as per usual definition) and at the same time ask if it "indeed diverges", I interpret the latter as a new term you implicitly introduce where "sum X indeed diverges" means "sum X can't have a value associated with it one way or another".

But if we interpret the question this way, no, you can't be sure: even if the product "indeed diverges" now, there is a chance that someone later will invent some other way to associate a value with it. Basically, predicate "X indeed diverges" is ill-defined.

Answer 2

For $n \ge 3$, $\ln n > 1$. So $\Pi_{n=3}^{+ \infty} \ln n$ is the product of positive numbers greater than 1 and hence it must diverge.