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 ?
 A: 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.
A: Let us construct a similar expression where we actually do know the exact answer. Let us say the series starting with the number $2$ having the result of every logarithm being equal to the number it is added to:
$$\ln(2+\ln(3.69 ...+\ln(20.11 ...+\ln(...))))$$
We can calculate each term in this expression as:
$$t_{1}=2,\ t_{n}=\frac{\exp(t_{n-1})}{2}$$
Since this is a super-exponential series and the gap between the first two terms is greater than $1$ it is safe to conclude that every term in the series is greater than the same term in the series of natural numbers.
As increasing any of the terms in the original expression can only cause the resulting value to be greater, this new expression can be used to calculate an upper bound for the original expression. We can in the original expression, for $n\ge2$ replace any $n+\ln(...)$ with $2n$, to get a result that is greater than the original expression, and thus a useful upper bound. Pair this with the lower bound of replacing with $n$ and you have a proof of convergence and a method of calculating arbitrarily precise upper and lower bounds.
