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This question is similar to one concerning nested logarithms and also nested radicals involving factorials. If I am correct a quick calculation shows:

$$\ln (1! \ln (2! \ln (3! \ln (4! \ln (5! \ln (6! \ldots ))))))\approx 0.654569$$

Does the sequence implied by this infinite nested logarithm converge to a finite value and if so to what value exactly?

Goldbug
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    Heuristically, the $\ln(x)$ function grows more slowly than $\sqrt{x}$ (in particular I believe $\ln(x)<\sqrt{x}$ for all $x>0$) so I would expect this to converge by the comparison test with the nested radical sequence, or something similar. – 79037662 Dec 19 '19 at 15:46
  • For clarification, you mean for "$1!(\ln 2!(\ln 3!(\cdots$" to all be in the argument of the first logarithm, including the items in the parentheses? At first glance, I would interpret $\ln a(b)$ to be $(\ln a)\cdot (b)$ where $b$ is not included as a part of the argument of the logarithm. – JMoravitz Dec 19 '19 at 16:06
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    The sequence seems to converge very rapidly. For $n=100$ and $n=200$ , the value differs less than $10^{-200}$. The limit is apparently $$0.65456942883021402957745237107331947\cdots $$ With pari/gp, we can calculate the value with this function : f(n)={m=n;s=m!;while(m>1,m=m-1;s=log(s)*m!);s=log(s);s} – Peter Dec 19 '19 at 16:06
  • @JMoravitz Yes, sorry my parenthesis were confusing, I edited them to be more clear. I mean for the logarithms to be nested within one another. – Goldbug Dec 19 '19 at 16:08
  • @Peter Yes, this matches almost exactly what I am getting with pari/gp up to 2000!. It does appear to converge very quickly... – Goldbug Dec 19 '19 at 16:11
  • @Goldbug If you increase the precision, you should get exactly the same decimal expansion upto this point. – Peter Dec 19 '19 at 16:47
  • But doesn't the numerical calculation rather calculate the sequence $(f_i(i)){i\in\mathbb{N}}$, for $f_k(n):=\begin{cases} \ln(n! f(n+1))&, 0\le n<k\k! &, n=k \end{cases}$, which assumes that $k!f\infty(k+1)\approx k!$ for $k\to\infty$ ? – Sudix Dec 19 '19 at 17:01
  • @Sudix I was also a little worried about this as it seems we are ignoring some large numbers by truncating the nesting? This seems similar to how the numerical calculation was done here and the answer seemed to come out correct however... – Goldbug Dec 20 '19 at 18:35
  • They all just calculate the limit of the truncated sequence in that answer. Nobody even touches the question of whether the initial expression actually is a sensible mathematical expression. – Sudix Dec 21 '19 at 16:14

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Consider $$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\dots}}}}$$ We have that $n!^{2^{-x}} \to 1$, which is sufficient to conclude the above converges. In other words, we have that $$f(1!f(2!f(3!f(\dots))))$$ Where $f(x)=\sqrt{x}$, converges. Noting that $g(x)=\ln(x)$ is positive but less than $f(x)$ for all $x>0$, it follows that $$g(1!g(2!g(3!g(\dots))))$$ also converges. The notation is a little icky, but you could also define it recursively. I checked the number computed by user Peter in the comments in OEIS and through an inverse symbolic calculator, neither of which turned up any results. If you wanted a closed form for what it converges to, you are not likely to find one, though its convergence is rapid and decimal approximations become very accurate very quickly.

Edit: slight mistake. That $n!^{2^{-x}}$ converges is sufficient to conclude $$\sqrt{1 + \sqrt{2! + \sqrt{3! + \dots}}}$$ converges, but it is not sufficient to conclude $$\sqrt{1!\sqrt{2!\sqrt{\dots}}}$$ converges. Determining whether the above converges is synonymous to showing $$\prod_{n=1}^\infty n!^{2^{-x}}$$ Converges. I will not show this, but you can take my word that it does converge. So, because the above converges, we have that $\sqrt{1\sqrt{2\dots}}$ converges. Because $\sqrt{x} > \ln(x)$, your infinitely nested logarithm must be bounded by $\sqrt{1!\sqrt{2!\dots}}$, and therefore must also converge.

Descartes Before the Horse
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