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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 $f_n \sim \dfrac{3n \ln(2)\ln(n)}{2}$.

TMM
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mick
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  • How much better do you need it? Apparently you have the leading order already figured out, as well as the accompanying leading order constant. – TMM May 13 '13 at 18:15
  • Well for instance a term like $\sqrt ln(ln(n))$ would be nice. – mick May 13 '13 at 18:22
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    Could you provide some details about your calculation? Maybe someone will be able to develop your idea and could derive more terms. – vesszabo May 13 '13 at 18:22
  • @vesszabo even that calculation is not formal. Handwaving stuff. Since it cannot be of the form n^a for a>1 and it must grow faster than O(n) we get C n ln(n) because of the finiteness of elementary functions. Not very formal hmm. I basically used elimination. – mick May 13 '13 at 19:31
  • $f_n / n\log(n)$ doesn't appear to be constant asymptotically, suggesting that the form you arrived at could indeed be improved. – Nathaniel Bubis May 13 '13 at 19:45
  • The analogous continuos ODE $f' = \ln f$ leads to $f = \exp ( E_i^{-1} [(c+x) e^c ]-c)$ where $E_i^{-1}$ is the inverse of the exponential integral function. But its asymptotics are ugly, see answer and linked paper: http://math.stackexchange.com/questions/94860/inverse-function-of-operatornamelix-over-x-mu – leonbloy May 15 '13 at 02:17
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    @leonbloy your ODE has the main solution : functional inverse of $li(x)$. Since $li(x)$ is well approximated by $\dfrac{x}{\ln(x)}$ , the inverse is about $x \ln(x)$. This suggests the main term for $f_n$ is $n \ln(n)$. I assume the difference between that main term and $ f_n$ is of the form $(A \cdot n \ln(n))^B$ where $B$ is probably $\frac{1}{2}$. If you disagree plz explain. Note however that even the ODE is just an approximation , so we need to be carefull with approximating an approximation ! – mick May 26 '13 at 23:35
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    Inverse of Li does seem to be the proper asymptotic. Indeed, $f_{20000} = 224080.77$ and $\mathrm{Li}(224080.77) = 19999.6$. And first-order for the inverse of Li is $n \log n$, constant 1, as Dld says. – GEdgar May 31 '13 at 15:31

1 Answers1

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$$\lim_{n\to\infty}\left(\frac{f_n}n-\log n-\log\log n\right)=-1$$

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.

Did
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  • I confess I haven't checked all the steps, but it looks nice. – leonbloy May 31 '13 at 13:38
  • Interestingly, this result does not agree with the OP lead term coefficient (though the difference is numerically small $1.0397$ instead of $1$), and it agrees with what I had got with a ODE approximation (see my comment in the question; I didn't post it because it lacked rigour, and only gave the first asymptotic term) – leonbloy May 31 '13 at 13:41
  • @leonbloy Thanks. Regarding your confession... ego te absolvo a peccatis tuis. Amen. – Did May 31 '13 at 17:03
  • Could you show more detailed steps plz ? – mick Jun 01 '13 at 21:59
  • Could you ask more focused questions in comments, thanks? By the way, here is a focused question: where did you find the factor $3\ln(2)/2$ in your post? – Did Jun 02 '13 at 14:08
  • @Did that was a mistake. It seemed to fit for small values and I jumped into conclusions... almost since it was a question NOT an answer :) – mick Jun 17 '13 at 21:47
  • I forgot to upvote it ! Did it today :) – mick Mar 13 '22 at 13:46