What is a good asymptotic for $f_n = f_{n-1}+\ln(f_{n-1})$? Let $f_0=2$ and $f_n=f_{n-1}+\ln(f_{n-1})$. What is a good asymptotic to the sequence $f_n$? With good I mean much better than $f_n \sim \dfrac{3n \ln(2)\ln(n)}{2}$.
 A: 
$$\lim_{n\to\infty}\left(\frac{f_n}n-\log n-\log\log n\right)=-1$$

To prove this, consider $g_n=n\log n+n\log\log n-n$ and $h_n=f_n-g_n$. One wants to prove that $h_n=o(n)$. The identity $f_{n+1}=f_n+\log f_n$ is equivalent to
$$
h_{n+1}=g_n+h_n+\log(g_n+h_n)-g_{n+1}.
$$
Using simple properties of the logarithm, one can show that this implies
$$
h_{n+1}=h_n+\log\left(1+\frac{\log\log n}{\log n}-\frac1{\log n}+\frac{h_n}{n\log n}\right)+O\left(\frac1{\log n}\right).
$$
In particular, if $h_n=o(n\log n)$, the logarithm in the RHS goes to zero hence $h_{n+1}=h_n+o(1)$, which implies $h_n=o(n)$. Thus, our task is to prove the easier statement that
$$
f_n=n\log n+o(n\log n).
$$
To do so, first note that $f_n\geqslant2$ for every $n\geqslant0$ yields $f_{n+1}-f_n\geqslant\log2$ hence $f_n\geqslant n\log2$ for every $n\geqslant0$. Plugging this once again in the recursion $f_{n+1}=f_n+\log f_n$ yields $f_{n+1}-f_n\geqslant\log n+\log\log2$ hence, summing up, $f_n\geqslant n\log n+o(n\log n)$. 
In the other direction, $f_{k+1}-f_k=\log f_k\leqslant\log f_n$ for every $k\leqslant n$ hence $f_n\leqslant f_0+n\log f_n$, which can be seen to imply $f_n\leqslant n\log n+2n\log\log n$ for every $n$ large enough. This completes the proof.
