Let's consider the recursive sequence: $$ x_0:=1\\ x_{n+1}:=1+e^{-e^{x_n}}. $$ Is it convergent or divergent?
What I have tried so far is the following:
For each member we can see that $1<x_n<2$. As the function $f(x)=1+e^{-e^{x}}$ is continuous, a limit $l$ (if it existed), must satisfy the equation $l=1+e^{-e^{l}}$. At first glance this seems impossible which would prove the divergence. However, if we define $h:(1,2)\to\mathbb{R}$, where $h(l)=l-1-e^{-e^{l}}$, which is a continuous function, we see that $h(1.5)>0$ and $h(1.0001)<0$. So by intermediate value theorem there must exist an element $l_0$ such that $h(l_0)=0$ which implies $l_0=1+e^{-e^{l_0}}$. So it seems that this approach is a dead end.
Maybe the sequence does converge? But it doesn't seem that it is monotone.
Do you have any tips which way to go?
Consider the function $u: (0,\infty)\to (0,\infty)$ defined by $u(t)=te^{-t}$. Because $u'(t)=(1-t)e^{-t}$, $u$ is increasing on $(0,1)$ and decreasing on $(1,\infty)$. Furthermore, $u$ has a maximum at $t=1$, in other words $u(t)\le u(1)=e^{-1}$ for all $t>0$.
Let us now consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by $$ f(t)=1+\exp(-e^t) $$ Since $f'(t)=-e^t\exp(-e^t)$, we have $$ |f'(t)|=e^t\exp(-e^t)=u(e^t) \le e^{-1} $$ It follows that $$ |f(x)-f(y)|=\left|\int_x^yf'(t)dt\right|\le e^{-1}|x-y| \quad \forall x, y \in \mathbb{R}, $$ i.e. $f$ is a contraction because $r=e^{-1}<1$.
For every integer $n\ge 1$ we have
\begin{eqnarray} |x_n-x_{n-1}|&=&|f(x_{n-1})-f(x_{n-2})|\cr &\le& r|x_{n-1}-x_{n-2}|\cr &\le&r^{2}|x_{n-2}-x_{n-3}|\cr &\vdots&\cr &\le&r^{n-1}|x_1-x_0| =\end{eqnarray}
For every integer $m,n$, with $m<n$, we have
\begin{eqnarray} |x_m-x_n|&=&|(x_{m}-x_{m+1})+(x_{m+1}-x_{m+2})+\ldots+(x_{n-2}-x_{n-1})+(x_{n-1}-x_n)|\cr &\le&|x_{m}-x_{m+1}|+|x_{m+1}-x_{m+2}|+\ldots+|x_{n-2}-x_{n-1}|+|x_{n-1}-x_n|\cr &\le&r^{m-1}|x_1-x_0|+r^{m}|x_1-x_0|+\ldots+r^{n-3}|x_1-x_0|+r^{n-2}|x_1-x_0|\cr &\le&(r^{m-1}+r^{m}+\ldots+r^{n-3}+r^{n-2})|x_1-x_0|\cr &=&r^{m-1}(1+r+r^2+\ldots+r^{n-m-1})|x_1-x_0|\cr &=&r^{m-1}\frac{1-r^{n-m}}{1-r}|x_1-x_0| \end{eqnarray} i.e. $$ |x_m-x_n|\le \frac{r^{m-1}-r^{n-1}}{1-r}|x_1-x_0| \quad \forall n>m $$
Since $r<1$, we have
$$ \lim_{m,n\to \infty}\frac{r^{m-1}-r^{n-1}}{1-r}|x_1-x_0|=0, $$ $\{x_n\}$ is a Cauchy sequence, and therefore is convergent. Furthermore, its limit is a fixed point of $f$