The limit of $\ln(1+\ln(2+\ln(3+...+\ln(n)))...)$

Does this limit: $$\lim_{n\to\infty}\ln(1+\ln(2+\ln(3+...+\ln(n)))...)$$

exist ? And if yes, which value does it have ?

Solution 1:

You can get a proof of convergence along these lines: Show by induction on $k$ that, for all $n$, $$\ln(n+\ln(n+1+\ln(n+2+\cdots+\ln(n+k))))\le\sum_{j=0}^k\frac{n!}{(n+j)!}\ln(n+j).$$ The basic inequality you will need in the induction step has the form $$\ln(n+a)<\ln n+\frac an,$$ which you apply with $a=\ln(n+1+\ln(n+2+\cdots+\ln(n+k)))$.

So, for $n=1$ you get $$\ln(1+\ln(2+\ln(3+\cdots+\ln(k+1))))\le\sum_{j=0}^k\frac{1}{j!}\ln(j),$$ and the series on the right clearly converges. The sequence on the left, on the other hand, is increasing, so boundedness implies convergence.

Solution 2:

Pari/GP:

n=28
res=0
forstep(x=n,1,-1,res =log(x+res))
res
 %553 = 0.820359862208789788473466794941

It does not change in the visible digits if $n$ is furtherly increased...

 t_n =res \\ save the result which we just got in a fixed variable
          \\ perform iteration on the inverse function
 x = t_n;for(k=1,n+3, x = exp(x)-k; print(x));

The result is:

1.27131705165
1.56554547884
1.78528449551
1.96127566254
2.10838918962
2.23496562706
2.34616059168
2.44538852642
...
3.41221000321
3.33220451018
-1.19274014861 E-170   \\ this is at iteration 28
-28.0000000000   \\ with small fractional parts exp(-k)
-30.0000000000   \\ ... 
-31.0000000000

So if we assume some limit $\lim_{n \to \infty} t_n$ then with an appropriate $\delta_m$ if we iterate from $$x_0=t_\infty+\delta_m \\ x_k=\exp(x_{k-1})-k $$ some $x_m=0$, $x_{m+1}=-m$ occurs...