if this limit $\displaystyle\lim_{n\to\infty}a_{n}$ exists, then find this $w$ Let the sequence $\{a_{n}\}$ be such that $$a_{0}=w,\quad a_{n+1}=w^{a_{n}},\quad w>0.$$
If $\displaystyle\lim_{n\to\infty}a_{n}$ exists, then find the possible values for $w$.
My try: assume that
 $\displaystyle\lim_{n\to\infty}a_{n}$ exists, and let
 $\displaystyle\lim_{n\to\infty}a_{n}=a$. 
Then
$$a=w^a$$@
 can prove when $w\in(\dfrac{1}{e},e^{\frac{1}{e}})$
But $w\in(0,\dfrac{1}{e})$.I can't
then I don't know how to proceed.
 A: This is an interesting question. In fact, for these sort of sequences that are generated by repeat iteration of a strictly monotonic function. There is something generic we can say.
Please pardon me for this lengthy answer.
Let $f: [0,\infty) \to [0,\infty)$ be any strictly increasing continous function on $[0,\infty)$ with at most countably many fixed points $x_1, x_2, x_3, \ldots \in (0,\infty)$,
all of them are isolated. Let $x_0 = 0$ and if the number of fixed points $N_f$ is finite, let $x_{N_f+1}$ be $\infty$. 
We can break $[0,\infty]$ into at most countably many intervals $[x_k, x_{k+1}]$. 
For any $b \in (x_k, x_{k+1})$, we have:
$$\begin{align} x_k & \begin{cases}
= 0 \;\;&\implies f(x) \ge x_0 = 0\\
\ne 0 &\implies x_k \text{ is a fixed point and } x_k < b \implies x_k = f(x_k) < f(b)
\end{cases}\\
\\
x_{k+1} & \begin{cases}
= \infty &\implies f(x) \le x_{k+1} = \infty\\
\ne \infty &\implies x_{k+1} \text{ is a fixed point and } b < x_{k+1} \implies f(b) < f(x_{k+1}) = x_{k+1}\end{cases}
\end{align}$$
i.e $f$ maps the intervals $[x_{k}, x_{k+1}]$ into themselves. If we construct a sequence
out of $b$ by repeat iteration of $f$:
$$b_0 = b,\; b_1 = f(b_0),\;b_2 = f(b_2),\; \ldots,\; b_{n+1} = f(b_n),\; \ldots$$
We have $b \in [x_k, x_{k+1} ] \implies b_n \in [x_k, x_{k+1}$ for all $n$. Furthermore,
it is easy to see
$$b_1 \begin{cases}
> b_0 &\implies b_2 = f(b_1) > f(b_0) = b_1 \implies b_3 > b_2 \implies \ldots\\
= b_0 &\implies b_2 = f(b_1) = f(b_0) = b_1 \implies b_3 = b_2 \implies \ldots\\
< b_0 &\implies b_2 = f(b_1) < f(b_0) = b_1 \implies b_3 < b_2 \implies \ldots
\end{cases}$$
This implies


*

*If $f(b) < b$, then $b_n$ will be a strictly decreasing sequence
bounded below by $x_k$. As a result, $b_{\infty} = \lim\limits_{n\to\infty} b_n$ exists.
Since $f$ is continuous, $b_{\infty}$ is a fixed point and hence must equal to $x_k$.

*If $f(b) > b$, then $b_n$ will be a strictly increasing sequence.
If $x_{k+1} < \infty$, then $b_n$ is bounded above by $x_{k+1}$ and by a similar argument
as above, we find $b_{\infty} = \lim\limits_{n\to\infty} b_n$ exists and equal to $x_{k+1}$.
If $x_{k+1} = x_{N_f+1} = \infty$, the sequence $b_n$ cannot be bounded, otherwise its limit $b_{\infty}$ will be a fixed point of $x = f(x)$ bigger than $x_{N_f}$, contradict with our definition of $N_f$. In both cases, we have $b_n \to x_{k+1}$.


What this means is if we generate a sequence $b_n$ by repeat iteration of a strictly increasing function start from a $b \in [x_{k},x_{k+1}]$, the corresponding limit $b_{\infty}$ always exists in $[0,\infty]$
and is given by
$$b_\infty \stackrel{def}{=} \lim_{n\to\infty} b_n = \begin{cases} 
x_{k},   &\text{ if } f(b) < b\\
x_{k+1}, &\text{ if } f(b) > b\\
b,       &\text{ if } b = x_{k} \text{ or } x_{k+1}
\end{cases}$$
Let's get back to our original problem. The discussion will be cleaner if we start the
sequence from $b = 1$ instead. Let $f(x) = w^x$ and $b_n$ be the sequence
$$b_0 = b = 1,\;b_1 = f(b) = w = a_0,\; b_2 = f(b_1) = f(a_0) = a_1,\;\ldots$$
When $w > 1$, $f(b) = w > 1 = b$ and $f$ is strictly increasing over $[0,\infty)$.
We can apply the result derived above. The fixed points of $x = f(x)$ is given by
$$x = w^x \quad\iff\quad x^{\frac{1}{x}} = w$$
If one make a plot of $y = x^{\frac{1}{x}}$ for $x$ varies over $[0,\infty)$, one will find


*

*When $w > e^{\frac{1}{e}}$, the line $y = w$ doesn't intersect with the curve $y = x^{\frac{1}{x}}$. $x = f(x)$ doesn't have any fixed points and $N_f = 0$. $$b = 1 \in [ x_0, x_1 ] = [0,\infty] \quad\implies\quad b_\infty = x_1 = \infty$$
i.e. the sequence $a_n$ diverges to $\infty$.

*When $w = e^{\frac{1}{e}}$, the line $y = w$ touches the curve $y = x^{\frac{1}{x}}$ at $(x,y) = (e, e^{\frac{1}{e}})$. $x = f(x)$ has only one fixed point at $e$. 
$$b = 1 \in [0,x_1] = [0,e] \quad\implies\quad b_\infty = x_1 = e$$
i.e. the sequence $a_n$ converges to $e$.

*When $1 < w < e^{\frac{1}{e}}$, the line $y = w$ intersect the curve $y = x^{\frac{1}{x}}$ at two places. There are two fixed points $x_1$ and $x_2$ satisfying
$1 < x_1 < e < x_2$. The smaller one can be expressed in terms of the Lambert's W function:
$$x_1 = \frac{W(-\log w)}{-\log w}\tag{*1}$$
Since $b = 1 \in [0,x_1]$, we have $a_n$ converges to $x_1$ again.


When $w < 1$, the situation is more complicated. The function $f(x) = w^x$ is no longer strictly increasing. Instead, it is strictly decreasing. By direct computation, one can verify
$$b_1 = w < b_2 = w^w < b_0 = 1\quad\implies\quad b_2 = f(b_1) > b_3 = f(b_2) > b_1 = f(b_0)$$
Combine them give us
$$b_1 < b_3 < b_2 < b_0$$
Since $f$ is strictly decreasing, $\tilde{f} = f\circ f$ is strictly increasing. Let $[\tilde{x}_k,\tilde{x}_{k+1}]$ and $[\tilde{x}_l,\tilde{x}_{l+1}]$ be the intervals
induced by fixed points $\tilde{f}$ where $1$ and $w$ belongs to. What we have developed so far tell us


*

*$b_2 = \tilde{f}(b_0) < b_0 \quad\implies\quad b_{\infty}^e \stackrel{def}{=} \lim_{n\to\infty} b_{2n}$ exists and equal to $\tilde{x}_k$.

*$b_3 = \tilde{f}(b_1) > b_1 \quad\implies\quad b_{\infty}^o \stackrel{def}{=} \lim_{n\to\infty} b_{2n+1}$ exists and equal to $\tilde{x}_{l+1}$.


As a consequence, the sequence $b_n$ converges if and only if 
$$b_{\infty}^e = b_{\infty}^o
\;\;\iff\;\; \tilde{x}_{k} = \tilde{x}_{l+1}
\;\;\iff\;\; \tilde{f} = f\circ f \text{ has an unique fixed point in }(w,1).
$$
When we look at the plot of $y = x^{\frac{1}{x}}$ again, one will find that for $w < 1$,
it has a unique fixed point $x_1 \in (0,1)$. Once again, it can be parametrized by the
formula in $(*1)$. Another plot of $x_1$ v.s. $w$ shows that $x_1 \in (w,1)$. This 
automatically give us at least one fixed point of $\tilde{f}$ over $(w,1)$. The question becomes whether there are others.
We will show that for $w \in [ e^{-e}, 1)$, the $x_1$ so constructed is the unique fixed point of $\tilde{f}$ when $x$ varies over $(0,1)$. Let $\varphi(x) = x + w^x$ and notice
$$\tilde{f}'(x) =  w^{\varphi(x)} \log(w)^2\quad\text{ and }\quad\varphi'(x) = 1 + \log w \cdot w^x$$
There are 3 cases we need to analysis:


*

*When $w \in [\frac{1}{e},1)$, we have $0 \le -\log w \le 1 \implies \varphi'(x) \ge 0$ over $[0,1]$. This means $\varphi(x)$ is monotonic increasing over $[0,1]$ and hence 
$\tilde{f}'(x) \le w^{\varphi(0)} \log(w)^2 < 1$. 

*When $w \in (e^{-e},1)$, $\varphi'(x) < 0$ for $x \sim 0$ and $\varphi(x)$ is decreasing there. One can check that $\varphi(x)$ reaches its minimum at some $x_{min}$
given by:
$$\varphi'(x_{min}) = 0 
\;\;\implies\;\; x_{min} = \frac{\log\left(-\frac{1}{\log w}\right)}{\log w}
\;\;\implies\;\;
\begin{cases} 
w^{x_{min}} &= -\frac{1}{\log w}\\
w^{w^{x_{min}}} &= e^{-1}
\end{cases} 
$$
With this, we can bound $\tilde{f}'(x)$ from above as
$$\tilde{f}'(x) < w^{\varphi(x_{min})} \log(w)^2 = -\frac{\log w}{e} < 1$$ 

*When $w = e^{-e}$, we have similar result as 2. Namely, $\tilde{f}'(x) \le 1$ and the
equality is achieved at and only at $x = x_{min} = \frac{1}{e} = x_1$.
In all 3 cases, we have $\tilde{f}'(x) < 1$ for $x \in (0,1) \setminus \{ x_1 \}$.
Mean value theorem then tell us it is impossible to have another fixed point of $\tilde{f}$
over such a range.
Combine all these analysis, we can conclude

For $w \in [ e^{-e}, e^{\frac{1}{e}} ]$, the sequence 
  $a_n$ converges to the fixed point $x_1$ given by formula $(*1)$.

To see what happens when one decreases $w$ below $e^{-e}$, one just need to plot $\tilde{f}(x)$ near $x_1$ for $w \sim e^{-e}$. Even though we still have only one fixed point for $f(x)$ at $x = x_1$. The fixed point of $\tilde{f}(x)$ now split into three when
$w$ is decreased below $e^{-e}$.  In terms of $a_n$, the sequence bifurate. The even and odd sub-sequences start converging to a different limit.
