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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

  1. 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

$${\sqrt{2}}^{{\sqrt{2}}^{{\sqrt{2}}^{{\sqrt{2}}^\cdots}}}=2 $$

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  • $\begingroup$ What about generalizing with $\exp\left(\exp\left(\sum_{i=1}^\infty \ln(\ln(a_i))\right)\right)$ ? $\endgroup$
    – user65203
    Jan 26, 2016 at 19:40
  • $\begingroup$ I thinkt that it is not correct. For only four "elements" we have $$a_1^{{a_2}^{{a_3}^{a_4}}}=x \Rightarrow a_4\ln{\left[a_3\ln{\left(a_2\ln{a_1}\right)}\right]}=\ln\ln\ln{x} $$ So this generalization shouldn't work :( $\endgroup$
    – TheKwiatek666
    Jan 26, 2016 at 21:09
  • $\begingroup$ I didn't say that this generalization is an exponential tower. It is a different thing, iterating the transform from product to sum. $\endgroup$
    – user65203
    Jan 26, 2016 at 22:17
  • $\begingroup$ I do not understand why you do not write $$ a_0 \\ \;_{a_1} a_0 \\ \;_{\;_{a_2}} \;_{a_1} a_0$$ and then in genereal $$ \;_{\;_{\vdots}} \;_{a_1} a_0$$ because we begin with the first coefficient $e_0=a_0$ and assign a base $a_1$ to it $e_1=a_1^{e_0}$ then $e_2=a_2^{e_1}$ and so on $\endgroup$
    – Gottfried Helms
    Jan 27, 2016 at 20:37
  • $\begingroup$ If we're talking about sequences of the form $s_n = a_n^{s_{n-1}}$ and $s_0 = a_0$. We know for sure that $s_n$ will converge if $\forall n$ $e^{-e}\leq a_n \leq e^{e^{-1}}$ $\endgroup$
    – Lee Fisher
    Jan 27, 2016 at 21:05

3 Answers 3

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I think you can find in WP's "tetration" the criterion for the powertower over the reals, which was established by L. Euler (but independently by others)

Over the complex numbers the range of convergence of the infinite tetration was determined by W. Thron(1957) "Convergence of Infinite Exponentials with Complex Elements" and based on his work improved by D. Shell(1961) "On the convergence of infinite exponentials" .
The article of D.Shell is online here ; the basic article of W.Thron is also online here at ams.

See also for an overview in the wiki of the tetration-forum but the Shell-article should be linked to in the WP-article too.

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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".

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  • $\begingroup$ I would say this should have been a comment, except that typesetting those formulas in a comment could be a bit awkward. $\endgroup$
    – David K
    Jan 26, 2016 at 19:36
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    $\begingroup$ My original version started with "this doesn't answer your question, but is too long for a comment...". Must stop revising myself! (you could also argue that I answered the question for one possible interpretation) $\endgroup$
    – user2469
    Jan 26, 2016 at 19:39
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I don't know any theorems for convergence of power towers, but here is food for thought.

Let's define pt(a, n) (pt as short for "power tower") of a sequence a as:

  • pt(a, 1) = a_1
  • pt(a, 2) = a_1^(a_2)
  • pt(a, 3) = a_1^(a_2^(a_3))
  • ...
  • pt(a, n) = a_1^(a_2^(a_3^...^(a_n)...))

For lim [n -> oo] pt(a, n) to exist, pt(a, n) should be very near to pt(a, n+1). This requires a_n^a_(n+1) to be very near to a_n, thus a_(n+1) very near 0.

So, a convergence criteria could be a_n nearing 0. Ironically, 0^0 (and 0^0^...^0) are indeterminate forms.

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