Does this pattern converge anywhere?

Question

Let $$L = \ln(2\ln(3\ln(4\ln(5\ln(6\ln(7 \cdots))))))$$

Then does $L$ converge to any finite value? If yes then how and which value does it converge to? If no then why?

While fiddling around with the calculator I saw that

1. $\ln(2\ln(3\ln(4\ln(5\ln(6\ln(7)))))) = 1.35280544$

2. $\ln(2\ln(3\ln(4\ln(5\ln(6\ln(7\ln(8))))))) = 1.3633034$

3. $\ln(2\ln(3\ln(4\ln(5\ln(6\ln(7\ln(8\ln(9)))))))) = 1.366565132$

So, does this converge anywhere?

Answer 1

Let $L_n= \ln(2\ln(3\ln(4\ldots \ln(n)\ldots )))$
Starting from "$n\ln(n+1)\geq n$" , it's not difficult to show that $(L_n)_{n\geqslant 2}$ is increasing.

We can show that $\ \forall x \in [1,+\infty[ \ , \ \ln(x) \leq \sqrt{x}$

Then: $\forall n \geqslant 4 \ , \ L_n \leq \sqrt{2\sqrt{3\sqrt{4 ...\sqrt{n}}}} = \exp\left( \displaystyle \sum_{k=2}^n \dfrac{\ln(k)}{2^{k-1}}\right) \leq\exp\left( \displaystyle \sum_{k=2}^{+\infty} \dfrac{\ln(k)}{2^{k-1}}\right) $

So, $(L_n)_{n\geqslant 2}$ is bounded and increasing. It's convergent.