How can I prove $x_{n+1} = e^{-x_n}$ is convergent? I'm doing a practice problem which asks to prove that the sequence defined by $x_{n+1} = e^{-x_n}$ is convergent (or rather "study the convergence of $(x_n)$"). So I'd like to try and find some sufficient condition on $x_0$ for the sequence to converge.
I can see that $e^{-x}$ is $k$-lipschitzian with $k<1$ on $[a, \infty[$ for $a>0$. But the problem is that $e^{-x}$ does not map $[a, \infty[$ into itself. I started trying to find values of $a, b$ such that $[e^{-b}, e^{-a}]\subseteq[a, b]$, but then I wondered if maybe there was some simpler approach that I was missing.
 A: The solution is a corollary of another Banach fixed point theorem: see here
Essentially, if we have contraction with constant $k=1$ and compactness (instead of completeness) the result will follow. 
For any $x_0$ we have $x_1 \geq 0$ and so $x_2 \leq e^0 = 1$. A trivial induction shows that $x_n \in [0,1]$ for $n\geq 2$
As @Jack M has pointed out in the comments, $e^{-x}$ is indeed 1-lipschitz on $[0,1]$ so the conditions of the theorem in the link I provided are met: $K = [0,1]$. Thus, the result follows.  
A: I like Milen's answer and upvoted it. Here's an approach that uses the standard contraction mapping theorem.
Let $f(x)=\exp(-x)$ and let $F(x)=f(f(x))$. We compute
\begin{align}
 F'(x) &= e^{-x-e^{-x}} \\
 F''(x) &= e^{-x-e^{-x}} \left(e^{-x}-1\right).
\end{align}
Note that $F''(x)<0$ for $x>0$. Thus $F'$ is decreasing on $[0,\infty)$ and
$$F'(x)<F'(0)=\frac{1}{e}<1$$
for $x\in[0,\infty)$. It follows that $F$ is a contraction on $[0,\infty)$ and, thus, has a unique fixed point in $[0,\infty)$, say $x_0$. I claim that your recursively defined sequence converges to $x_0$.
Let $a$ be a real number with $0<a<x_0$ and let $b=f(a)$. Our observations on $F$ imply that $a<f(b)<x_0<f(a)<b$. In particular, $f$ maps $[a,b]$ into itself. Furthermore, it's easy to see that $|f'(x)|<|f'(a)|<1$ on $[a,b]$ so that $f$ is a contraction on $[a,b]$. Thus, iteration of $f$ must converge to $x_0$ when we start in $[a,b]$. Finally, every starting point in $\mathbb R$ must eventually land in some interval of the form $[a,f(a)]$.
