Nested logarithm sequence $a_{n+1}=a_n+\log a_n$, growth rate

Question

I have considered a sequence defined as follows:

$$a_{n+1}=a_n+\log a_n$$

$$a_1=2$$

Surprisingly enough, its growth is almost linear, or, more precisely, it seems that:

$$n < a_n < n^{3/2}$$

From numerical experiments, the growth rate for large $n$ is well described by a power law (as it's obvious from the previous condition), with the exponent being around $1.2$. Here's the plot of:

$$r_n=\frac{\log a_n}{\log n}$$

for $n=10^3 - 10^5$.

For larger $n$ we have, for example:

$$r_{0.5 \cdot 10^6}=1.204993305$$

$$r_{1.5 \cdot 10^6}=1.194589226$$

$$r_{4.5 \cdot 10^6}=1.185287843$$

$$r_{10 \cdot 10^6}=1.179122734$$

$$r_{20 \cdot 10^6}=1.174129607$$

Can we (1) prove that the sequence obeys a power law for $n \gg 1$ and (2) find the value of $r_{\infty}$?

Edit

If, as Did says in the comments, the sequence actually obeys another law:

$$a_n \asymp c n \log n$$

Then it would be interesting to prove and especially find $c$.

From experiments:

$$c_{10 \cdot 10^6}=1.113128774$$

$$c_{20 \cdot 10^6}=1.111029684$$

And the plot for $n=10^3 - 10^5$:

Professor Vector suggested that $c_{\infty}=1$, which is possible even though convergence is really slow.

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Update:

In a linked question I prove (kind of) that the following inequality holds for $n \geq 2$:

$$a_n \leq (n+1) \left( \ln (n+1) + \ln \ln (n+1) \right)$$

This bound is quite tight.

Answer 1

Hint. Note that if we show that eventually $$c_1 n\ln n\leq a_n \leq c_2 n\ln n\quad \text{with $0<c_1\leq c_2$}$$ ($c_1=1/2$ and $c_2=2$ should work by induction), then $\ln(a_n)\sim \ln(n)$, and, by Stolz-Cesaro Theorem, $$\lim_{n\to +\infty}\frac{a_n}{n\ln n}=\lim_{n\to +\infty}\frac{a_{n+1}-a_n}{(n+1)\ln (n+1)-n\ln(n)}= \lim_{n\to +\infty}\frac{\ln(a_n)}{\ln ((1+1/n)^n)+\ln(n+1)}=1.$$

Answer 2

It isn't a power law, and $\lim_{n\to\infty}r_n=1$. One way to prove this is to check by induction that $a_n\leq 2n\log n$ for every $n\geq 2$.