TITLE: Asymptotic behavior of iterative sequences
QUESTION [12 upvotes]: Suppose you have a sequence
$$
a_1<a_2<\cdots
$$
with
$$
a_{n+1}=a_n+f(a_n)
$$
where $f$ is a sufficiently nice nondecreasing function. What can be determined about the asymptotic behavior of $a_n$?
For example, suppose $f(x)=\log x+O(1)$ with $f(a_1)>0$. Can we conclude that $a_n\sim Cn\log n$ for some $C$?

REPLY [2 votes]: For the moment I will only provide a solution for this problem : 
Exercise. Given $x_0 >e^{-C}$ and $x_{n+1}=x_n+f(x_n)$ where $f(t):=\log t + C +o(1)$. Show that $x_n \sim n \log n$. 
The following proof has not been proofread yet so there might be some errors (and I hope no real errors), I will proofread when I have time. 
Proof. First $(x_n)$ is an increasing sequence and thus have a limit $\ell > e^{-C}$, but if it is finite then $\log \ell + C =0$ but it is not the case.
So for the moment we only have $x_n \to + \infty$. To be more precise one powerful method is the following : $x_{n+1}-x_n \simeq \log x_n$. The "continuous" problem associated is $y'(t) = \log y(t)$ that writes : $$\int_2^{y(t)} \frac{\mathrm{d}u}{\log u} = t- 2$$ hence we get $\frac{\mathrm{d}}{\mathrm{d}t} F(y(t)) = 1$ with : $$F(z)= \int_2^z \frac{\mathrm{d}u}{\log u}.$$ One can see (using by part integration) that : $$F(z) \sim \frac{z}{\log z}$$ as $z \to \infty$, we will use it shortly. 
Then it might be interesting to look at the discrete equivalent for $\frac{\mathrm{d}}{\mathrm{d}t} F(y(t)) = 1$ which is : $$\int_{x_n}^{x_{n+1}} \frac{\mathrm{d}u}{\log u} \to 1.$$ In fact we can achieve just integrating the comparaison : $$\log x_n \leqslant \log u \leqslant \log x_{n+1} = \log x_n +o(\log x_n).$$
Thus we have $\displaystyle \int_2^{x_n} \frac{\mathrm{d}u}{\log u} \sim n$ thus we get $\frac{x_n}{\log x_n} \sim n$ and so $\log x_n \sim \log n$ and finally $x_n \sim n \log n$.