Convergence of a Recursive Sequence A friend gave me the following problem:

Let $c$ be any positive real number.  Define a sequence recursively by
  $$a_0=c,\;\text{and }\; a_n=c^{a_{n-1}}\;\text{for }\;n=1,2,\ldots$$
  For what values of $c$ does this sequence converge?

The problem is trickier than it first seems, as I believe there are values $c>1$ for which the sequence converges, and also values $0<c<1$ for which the sequence diverges.  This is supposed to be able to be solved by a first year calculus student, so elementary methods are preferred.
A followup question: Suppose the answer to the first question is some set $D\subset\mathbb{R}^+$.  Then we have a well-defined function $f:D\rightarrow\mathbb{R}$ given by $f(c)=L_c$ where $L_c$ is the limit of the sequence defined above.  Is $f$ continuous?  Is $f$ differentiable?
Thanks!
 A: This operation is known as Tetration.
As shown in wikipedia, Euler proved that it converges for $e^{-e}\leq c\leq e^{1/e}$ and diverges otherwise.
A: Define $A=a_\infty$ Then $c^A=A$, with $A$ is a fixed point of $c^z$. Consider that for an infinitesimal $dz$ 
$c^{A+dz}=c^{dz} c^A =c^{dz} A = (1+ \ln(c){dz})A = A+ \ln(c^A){dz} = A+ \ln(A){dz}$. 
So exponentiation maps ${A+dz}$ into $A+ \ln(A){dz}$ or more simply ${dz}$ into $\ln(A){dz}$. But this means that $\ln(A)$ is the multiplier or Lyapunov characteristic at $A$ of the fixed point of $c^z$.    
The question now is where is the multiplier exactly on the unit circle. For multipliers inside the unit circle we have convergence and with the multiplier outside the unit circle we have divergence. Therefore $\ln(A) = exp(2 \pi i x)$ and $A = exp(exp(2 \pi i x))$ where $x$ ranges from zero to one. 
Since $a^A=A$, $A^\frac{1}{A} = a $. So the curve $exp(exp(2 \pi i x))^{exp(exp(2 \pi i x))^{-1}}$ gives the boundary of convergence.
See Projective Fractals.
