The question is simple. Does $$\prod _{n=2}^{\infty }\left(1+\frac{1}{n \, \ln(n)}\right)$$converge?
I know that it is pretty clear to show that
$$\prod _{n=1}^{\infty }\left(1+\frac{1}{n}\right)$$
diverges, while
$$\prod _{n=1}^{\infty }\left(1+\frac{1}{n^2}\right)$$
converges to $ \dfrac{\sinh(\pi)}{\pi} $.
Does it diverge in the way that $$\sum_{n=2}^{\infty} {\frac{1}{n \, \ln(n)}}$$ does?
How would you show this?
No, it does not converge. First note that an infinite product $\prod\limits_{n=2}^\infty a_n$ converges iff $\sum\limits_{n=2}^\infty\log(a_n)$ converges. Then, by the limit comparison test for $x$ and $\log(1+x)$ we have,
$$\sum\limits_{n=2}^\infty \log\left(1+\frac{1}{n\log n}\right)\sim\sum\limits_{n=2}^\infty\frac{1}{n\log n},$$
where $\sim$ denotes that either both series converge or both diverge. You have already stated that the right hand side series diverges, so the left hand side must as well.