sequence $x_{n+1}=x_n+e^{-2018x_n}$ Let the sequence $(x_n)_{n \geq 0}$ defined by 
\begin{cases}
x_0=2017 \\
x_{n+1}=x_n+e^{-2018x_n}
\end{cases}
for every $n \ge 0$. Calculate $$\lim_{n\to \infty}x_n \text{ and }  \lim_{n\to \infty}\frac{\ln(2018+n)}{x_n}$$
I proved that $\displaystyle \lim_{n\to \infty}x_n=\infty$, but I don't realise how to find the second limit. I tried Cesaro-Stolz lema, but it doesn't work. Please help me!
 A: $$x_{n+1}=x_n+e^{-2018x_n}$$ means
$$e^{2018x_n}\,(x_{n+1}-x_n)=1,$$ i.e.
$$\sum^{n-1}_{k=0}e^{2018x_k}\,(x_{k+1}-x_k)=n.\tag{crucial}$$ Since $e^{2018x}$ is monotone increasing, this means $$\int^{x_n}_{x_0}e^{2018x}\,dx\ge n,$$ implying
$$\frac{e^{2018x_n}-e^{2018\cdot2017}}{2018}\ge n.$$ This gives
$$e^{2018x_n}\ge2018\,n+e^{2018\cdot2017},$$
and thus 
$$x_n\ge\frac1{2018}\ln(2018\,n+e^{2018\cdot2017})\tag{lower}$$ and
$$x_{n+1}-x_n=e^{-2018x_n}\le\frac1{2018\,n+e^{2018\cdot2017}}.$$
This is one of the mean aspects of this problem: $x_n$ diverges, but very slowly.
Whatever, we know $$e^{2018\,x_{n+1}}\le e^{2018\,x_n}\cdot \exp\left(\frac{2018}{2018\,n+e^{2018\cdot2017}}\right),$$
that is $$e^{2018\,x_{n+1}}\le C\,e^{2018\,x_n},$$  where $C$ is a constant very near to (but greater than) $1$. But then, (crucial) implies
$$\sum^{n-1}_{k=0}e^{2018x_{k+1}}\,(x_{k+1}-x_k)\le C\,n,$$ so $$\int^{x_n}_{x_0}e^{2018x}\,dx=\frac{e^{2018x_n}-e^{2018\cdot2017}}{2018}\le C\,n,$$ leading to $$x_n\le\frac1{2018}\ln(2018\,C\,n+e^{2018\cdot2017})\tag{upper}$$
From (lower) and (upper), it's clear that $$\lim_{n\to \infty}\frac{\ln(2018+n)}{x_n}=2018.$$
