Infinite exponentials We can read a lot of about convergence of series or Infinite products.
E.g. for series.
Following series
$$\sum_{i=1}^\infty a_i$$
is convergent when 
$$\lim_{n\rightarrow\infty}a_n=0$$
and


*

*D'Alembert's criterion


$$\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|<1$$
2. Cauchy's criterion
$$\limsup_{n\rightarrow\infty}\sqrt[n]{\left|a_n\right|}<1$$
3. Raabe–Duhamel's criterion
$$\lim_{n\rightarrow\infty}n\left(\frac{a_{n+1}}{a_n}-1\right)>1$$
For products we can use above criterias for series.
Following product converges
$$\prod_{i=1}^\infty a_i$$
if series
$$\sum_{i=1}^{\infty} \ln{a_i}$$
converges.
What about Infinite exponentials? (I don't know how to write it)
$$a_1^{{a_2}^{{a_3}^{{a_4}^\cdots}}} $$
I know that tetration is particular case of that for 
$$
a_1=a_2=a_3=\cdots=x
$$
Then this infinite exponentials convergence for 
$$
x\in\left<\frac{1}{e^e};\sqrt[e]{e}\right>
$$
Is there general theorem of that? Where can I read about results?
I'm sure that $\lim_{n\rightarrow\infty}a_n=0$ or $\lim_{n\rightarrow\infty}a_n=1$ is not required. Because for $a_i=\sqrt{2}$ we have
$${\sqrt{2}}^{{\sqrt{2}}^{{\sqrt{2}}^{{\sqrt{2}}^\cdots}}}=2 $$

Clarification request: are you defining your series as: 
$s(n)=s(n-1)^{a(n)}$ 
If so, you'll just get 
$s(n)=a(1)^{\prod _{i=2}^n a(i)}$ 
Or are you defining it so that you replace a(n-1) with $a(n-1)^{a(n)}$ 
in the previous term? 
In other words, 
$s(n)=\left(s(n-1)^{\frac{1}{a(n-1)}}\right)^{(a(n-1)^{a(n)})}$ 
It looks like you're doing the latter (which is the harder case and 
really doesn't look much like a series), but I wanted to doublecheck. 
A more natural exponential "series" might be:
$s(n)=a(n)^{s(n-1)}$
although the powers would then come out "backwards".
