Does the infinite nested logarithm $\ln(2\ln(3\ln(4\ln(5\ln(6...)))))$ converge?

I was playing around with nested radicals and I decided to see if nested equations of logarithms would converge.

It seems to converge to a value around $1.368$, and at a depth of 20 it has a value of $1.3679012...$, however I am not sure how to prove whether it actually does converge.

Solution 1:

For all integers $n \geq 2$, note that $$n(n+1) < e^n. \tag{*}$$ We may show this by expanding $e^n > 1 + n + \frac{n^2}{2} + \frac{n^3}{6}$, which implies: $$e^n- n(n+1) > 1 - \frac{n^2}{2} + \frac{n^3}{6} = \left(\frac{n}{6} - \frac{1}{2}\right)n^2 + 1.$$ The RHS is manifestly positive for $n \geq 3$ and can be checked to be positive for $n = 2$.

Now, starting with $\ln n < n$ (which should need no proof), multiply on both sides by $n-1$ and apply $(\text{*})$ to show that $$(n-1) \ln n < (n-1) n < e^{n-1}.$$ Take logarithms to get $$\ln ( (n-1) \ln n) < n-1,$$ then multiply on both sides by $n-2$ to show that $$(n-2) \ln ( (n-1) \ln n) < (n-2) (n-1) < e^{n-2},$$ or $$\ln ((n-2) \ln ( (n-1) \ln n)) < n-2.$$ Multiply by $n-3$, apply $(\text{*})$, and take logarithms again to get $$\ln ((n-3) \ln ((n-2) \ln ( (n-1) \ln n))) < n-3.$$ By proceeding similarly, we can show that for $2 \leq k < n$ arbitrary, $$\ln (k \ln ((k+1) \cdots \ln n)) < k.$$ The LHS of this inequality is monotone increasing and bounded above as $n \to \infty$, so it must have a limit $$\ln (k \ln ((k+1) \ln ((k+2) \cdots))) \leq k.$$ In particular, $$\ln (2 \ln (3 \ln (4 \ln (5\cdots )))) \leq 2.$$

Solution 2:

For $n\le m$, let $$ f(n,m)=n\ln((n+1)\ln(\ldots (m)\ldots))$$ i.e., $$f(n,m)=\begin{cases}n&n=m\\n\ln(f(n+1,m))&n<m\end{cases} $$ We want $\lim_{m\to\infty}f(1,m)$.

Clearly, $f(n,\cdot)$ is increasing (in particular, $f(n,m)\ge n$) so that convergence equals boundedness. Compare $f(n+1,m+1)$ against $f(n,m)$. If $m=n>10$, $f(n+1,m+1)=f(n,m)+1<2f(n,m)$. By induction on $m-n$, for $m>n\ge 10$ as well $$ \begin{align}f(n+1,m+1)&=(n+1)\ln( f(n+2,m+1) )\\&<(n+1)\ln(2f(n+1,m))\\&=(n+1)(\ln 2+\ln(f(n+1,m))\\&<(1+\tfrac1{10})n\cdot (1+\tfrac{\ln2}{\ln11})\ln(f(n+1,m))\\&<2f(n,m)\end{align}$$ So $$f(n,m)<f(n+1,m+1)<2f(n,m)\qquad m\ge n\ge 10 $$ This makes $$\tag1f(n,m)<f(n,m+1)=n\ln f(n+1,m+1)<n\ln f(n,m)+n\ln 2 $$ for $n\ge 10$. The right hand side is slower than linear in $f(n,m)$, hence $f(n,m)$ is bounded from above, $\lim_{m\to\infty}f(n,m)$ exists and ultimately so does $\lim_{m\to\infty}f(1,m)$

Remark: Numerically, $(1)$ gives us $f(10,m)<44.998$. This trickles down to an upper bound $$f(1,m)< 1.36794$$ But similarly, we find $f(20,m)<107$ and with that can improve the bound to $$f(1,m)<1.3679012618$$ (for comparison, $f(1,20)>1.3679012615$). Starting with a bound for $f(50,m)$, we can compute $$\lim f(1,m)=1.367901261797085169668909175760\ldots$$ to 30 decimals.