sequences and series - Compute: $limlimits_{ntoinfty} frac{x_n}{ln {n}}$

Let be the sequence $(x_n)_{n\geq0}$, $x_0$ a real number, and defined as follows: $$x_{n+1} = x_n + e^{-x_n}$$
Compute the limit:

$$\lim_{n\to\infty} \frac{x_n}{\ln {n}}$$

Luckily, I've found another post here with a very similar question. If you know other ways to solve it and want to share them then I'd be very grateful for that.

Answer

Heuristically, this difference equation resembles the differential equation $x'(t) = e^{-x}$, which has solutions $x(t) = \ln(t + C)$, so I would expect the answer to be $1$.

In fact, let $x_n = \ln(n + s_n)$, and $y(t) = \ln(t+s_n)$. Then for $n \le t < n+1$ we have
$y(t) > y(n) = x_n$, so
$$\eqalign{ \ln(n+1+s_n) - \ln(n+s_n) &= \int_n^{n+1} \dfrac{dt}{t+s_n} = \int_n^{n+1} e^{-y(t)}\ dt\cr &< \int_n^{n+1} e^{-x_n}\ dt = e^{-x_n} = x_{n+1} - x_n\cr}$$

i.e. $x_{n+1} - \ln(n+1+s_n) > x_n - \ln(n+s_n) = 0$, or $s_{n+1} > s_n$.
Thus $x_n > \ln(n+s_0)$. But then $x_{n+1} - x_n = e^{-x_n} < \dfrac{1}{n + s_0}$
so $\displaystyle x_n < x_0 + \sum_{j=0}^{n-1} \frac{1}{j+s_0}$. Since both upper and lower bounds are $\ln(n) + O(1)$, we conclude that $x_n = \ln(n) + O(1)$.