# limit of a recursively defined function

I was just going through some of the problems from an intro analysis course I took in the fall 2010 and was hoping people could shed some light on some problems I feel that I never fully solved. Here's a problem in question:

Let $x_0$ be a real number, and define a sequence $(x_n)_{n\geq0}$ recursively by setting $$x_{n+1} = x_n + e^{-x_n} ,$$ for all $n\geq0$. Evaluate (with proof) $$\lim_{n\to\infty} (x_n - \ln(n+1)).$$

We solved several problems involving such recursively defined limits, and almost always employed Cesàro means, and I always struggled with them. If someone could sketch a proof (or give steps for me to fill in) I would be more than grateful. I've googled a little bit looking for similar examples to get more practice, but haven't had much luck so if anyone knows any other similar problems (this is about the only thing I've found), I'd be happy to hear them as well.

-

I guess you managed to prove that $x_n$ is increasing and $x_n \to \infty$.

Denote $y_n=e^{x_n-\ln(n+1)}=\displaystyle\frac{e^{x_n}}{n+1}$.

Now we attempt to use Stolz-Cesàro:

$$\frac{e^{x_{n+1}}-e^{x_n}}{n+2-(n+1)}=e^{x_n}(e^{x_{n+1}-x_n}-1)=\frac{e^{e^{-x_n}}-1}{e^{-x_n}}\to 1$$

since $\displaystyle \lim_{y\to 0}\frac{e^y-1}{y}=1$ and $e^{-x_n}\to 0$.

By Stolz-Cesàro, $y_n \to 1$ and this means that $\lim_{n \to \infty}x_n-\ln(n+1)= 0$.

-
+1. Very elegant! –  Hans Lundmark Aug 21 '11 at 21:26