(The presentation that follows is a very light introduction. For more austere results, consult Manuscript 1).

Let's begin with some notation. Let n in N be a positive integer. We define as usual:

n! = 1*2*3*...*(n-1)*n.

Now we will define a new sequence of iterated factorials for n in N, as follows:

hf(n,k) = {n! iff k=1, [hf(n,k-1)]!, iff k > 1}. (1)

hf(n,2) = [n!]!, hf(n,5) = [[[[n!]!]!]!]!, etc. Where no confusion shall arise, we will use hf(n,k) = n!!...!, (k-"!"s). Some examples with Maple:

3!! = 720

4!! = 620448401733239439360000

3!!! =~ .2601218944 * 10¹⁷⁴⁷

etc.

We also use the hyperpower function defined here as:

F(x,y,k) = {x^y iff k =1, x^F(x,k-1), iff k > 1}, so,

F(x,1,1) = x,

F(x,1,2) = ²x = x^x,

F(x,y,2) = x^xy

...

F(x,1,k) = ^kx = x^xx...x (k-x's).

First, some elementary facts.

The comparison property of divisibility:

a | b and b =/= 0, => |a| ≤ |b|. (1)

Proof:

In any standard Number Theory book.

Corollary:

a | b and a, b > 0, => a≤b.

A couple of immediate facts about the sequence hf(n,k):

Lemma #1:

If m and n are in N, then: m≤n <=> hf(m,k) ≤ hf(n,k), for all positive k.

Proof:

Trivial by induction on k. The factorial sequence is 1-1 and strictly increasing.

Lemma #2:

If m and n are in N, then m≤n <=> m! | n!

Proof:

If m≤n, clearly m! | n!. On the other hand, m! | n!, along with the corollary, => m! ≤n!, and the lemma follows from lemma #1 (for k=1).

Lemma #3:

If m and n are in N, then: m≤n <=> hf(m,k) | hf(n,k), for all k in N.

Proof:

By induction on k. For k=1, this is lemma #2. Assume that the lemma holds for k in N, k>1. That is, m≤n <=> hf(m, k) | hf(n,k). From the induction hypothesis, m≤n, => hf(m,k) | hf(n,k). From the corollary, => hf(m,k) ≤ hf(n,k). From lemma #2, => [hf(m,k)]! | [hf(n,k)]!, => hf(m,k+1) | hf(n, k+1).

For the reverse, assume hf(m,k+1) | hf(n, k+1). From the corollary, => hf(m,k+1) ≤ hf(n,k+1). From lemma #1 (for arbitrary k), => m≤n, and the lemma follows.

Let's now look how fast these hyperfactorials grow. We will need the following two lemmas along with their corollaries.

Lemma #4:

If k ≥ 1 then for all n in N, hf(n+1,k+1) - hf(n,k+1) > hf(n,k+1)*[n^{hf(n+1,k) - hf(n,k)} - 1]

Proof:

Directly or by induction. The direct proof is shorter. By lemma #3, hf(n,k+1) | hf(n+1,k+1), => hf(n+1,k+1) - hf(n,k+1) = 1*2*...*hf(n+1,k) - 1*2*...*hf(n,k) = 1*2*...*hf(n,k)*[(hf(n,k)+1)*(hf(n,k)+2)*...(hf(n+1,k)) - 1] = hf(n,k+1)*[(hf(n,k)+1)*(hf(n,k)+2)*...(hf(n+1,k)) - 1]. If k ≥1, => hf(n,k) + j > n, for all j. There are hf(n+1,k) - hf(n,k) terms in the product, and the lemma follows.

Corollary:

If k ≥1 then for all n in N, hf(n+1,k) - hf(n,k) > ^kn = F(n,1,k)

Sketch of Proof:

Iterate the weaker inequality of lemma #4. hf(n+1,k+1) - hf(n,k+1) > n^{hf(n+1,k) - hf(n,k)}, and use induction.

Lemma #5:

If k ≥1, then hf(n,k)/hf(n+1,k) < 1/^kn = 1/F(n,1,k)

Proof:

Using lemma #3, hf(n,k) | hf(n+1,k), => hf(n,k)/hf(n+1,k) =

Using the corollary above, the product contains at most ^k-1n terms, and hf(n,k-1) + j > n, for all j, => hf(n,k)/hf(n+1,k) < 1/n ^k-1n = 1/^kn and the lemma follows.

Lemma #6:

If k ≥1, then

converges for all x.

Proof:

If k=1, we get the exp series. If k>1 then |a(n+1)/a(n)| = |x*hf(n,k)/hf(n+1,k)|. By lemma #5, the above tends to 0, for all x.

Lemma #7:

If k =1 then the series,

converges for |x| < 1/e. If k > 1 then it converges for all x.

Proof:

If k=1, the Ratio test gives lim_n->+∞|a(n+1)/a(n)| = |ex| < 1, provided |x| < 1/e. >If k > 1, the same test gives |a(n+1)/a(n)| = (n+1)⁽ⁿ⁺¹⁾*x/[nⁿ*(n+1)] *(n+1)*hf(n,k)/hf(n+1,k)|. Using lemma #5, |a(n+1)/a(n)| < |(n+1)⁽ⁿ⁺¹⁾*x/[nⁿ*(n+1)] *(n+1)/^kn|

lim_n->+∞[(n+1)⁽ⁿ⁺¹⁾*x/[nⁿ*(n+1)]] = ex, as before, and obviously lim_n->+∞(n+1)/^kn = 0, for k > 1, and the rest of the lemma follows.

Can we up the stakes a bit? What about x^xn on the numerator of the series instead of xⁿ? Let's see.

Lemma #8:

If k > 2 then the series:

converges for all x.

Proof:

Using lemma #5 |a(n+1)/a(n)| < |x^xn*[x-1]/^kn|. For sufficiently large n, x < n, => |a(n+1)/a(n)| < |nⁿⁿ/^kn|=|³n/^kn|, so if k > 2, => lim_n->+∞|a(n+1)/a(n)| < 1, and the lemma follows.

Lemma #9:

If k > m - 1 then the series,

converges for all x.

Proof:

Similar to the proof of lemma #8.

Let's see if we can go further up. What about ⁿx on the numerator?

Lemma #10:

If k ≥1 then the series,

converges for x in (0,e^(1/e)].

Proof:

Using lemma #5, |a(n+1)/a(n)| < |ⁿ⁺¹x/ⁿx * 1/^kn|. When x is in [(1/e)^e,e^(1/e)], lim_n->+∞ⁿx = e^-W(-ln(x)), by the article on Infinite Exponentials, so, lim_n->+∞ⁿ⁺¹x/ⁿx = 1, while lim_n->+∞1/^kn = 0, for k ≥1, and the lemma follows.

If on the other hand, x is in (0,(1/e)^e), |a(n+1)/a(n)| = |ⁿ⁺¹x/ⁿx*hf(n,k)/hf(n+1,k)|, and ⁿx is a two cycle, bounded above by 1 and below by 0. This means that we may have problems with the odd subsequence: ^2m+1x, on the denominator. Using lemma #5, => |a(n+1)/a(n)| < |ⁿ⁺¹x/ⁿx*1/^kn|, as before, and lim_m->+∞^2mx = a < 1, with a being a solution to x^xa=a, (See Solving the Second Real Auxiliary Exponential Equation), lim_m->+∞^2m+1x = b > 0, b < a, and with b given by x^a = b, above. This means that |ⁿ⁺¹x/ⁿx|, n in N, is a two-cycle. We check the even and odd subsequences: lim_m->+∞|a(2m+1)/a(2m)| ≤ lim_m->+∞|^2m+1x/^2mx*1/^k(2m)| = |b/a*0| = 0, while, lim_m->+∞|a(2m+2)/a(2m+1)| ≤ lim_m->+∞|^2m+2x/^2m+1x*1/^k(2m+1)| = |a/b*0| = 0, and the lemma follows^[1].

Lemma #11

If k >1 then for all n in N, hf(n+1,k+1) - hf(n,k) > hf(n,k)*[n^{hf(n+1,k) - hf(n,k-1)} - 1]

Proof:

Directly or by induction. The direct proof is shorter. By lemma #3, hf(n,k) | hf(n+1,k), and obviously hf(n+1,k) | hf(n+1,k+1), => hf(n+1,k+1) - hf(n,k+1) = 1*2*...*hf(n+1,k) - 1*2*...*hf(n,k-1) = 1*2*...*hf(n,k-1)*[(hf(n,k-1)+1)*(hf(n,k-1)+2)*...(hf(n+1,k)) - 1] = hf(n,k)*[(hf(n,k-1)+1)*(hf(n,k-1)+2)*...(hf(n+1,k)) - 1]. If k >1, => hf(n,k-1) + j > n, for all j. There are hf(n+1,k) - hf(n,k-1) terms in the product, and the lemma follows.

Corollary:

If k >1 then for all n in N hf(n+1,k+1) - hf(n,k) > ^k+1n = F(n,1,k+1)

Proof

Iterate the weaker inequality of lemma #11, hf(n+1,k+1) - hf(n,k) > n^{hf(n+1,k) - hf(n,k-1)}, and use induction.

Lemma #12:

If k >1, then, hf(n,k)/hf(n+1,k+1) < 1/^k+1n = 1/F(n,1,k+1)

Proof:

By lemma #3, hf(n,k) | hf(n+1,k), and obviously hf(n+1,k) | hf(n+1,k+1), => hf(n,k)/hf(n+1,k+1) =

Using the corollary above, the product contains at most ^kn terms, and hf(n,k-1) + j > n, for all j, => hf(n,k)/hf(n+1,k)< 1/n ^kn = 1/^k+1n and the lemma follows.

We can still go further.

Lemma #13:

The series,

converges for all x > 0.

Proof:

Using lemma #12, |a(n+1)/a(n)| < |ⁿ⁺¹x/ⁿx*1/ⁿ⁺¹n|. If x is in [(1/e)^e,e^(1/e)] then lim_n->+∞ⁿx = e^-W(-ln(x)), so, => lim_n->+∞|ⁿ⁺¹x/ⁿx| = 1, as before and lim_n->+∞|1/ⁿ⁺¹n| = 0, and the result follows. If x > e^(1/e), for sufficiently large n, x < n, => ⁿ⁺¹x < ⁿ⁺¹n. Using lemma #12, => |a(n+1)/a(n)| <  |1/ⁿx|, consequently, lim_n->+∞|a(n+1)/a(n)| = 0, and the result follows.

If on the other hand, x is in (0,(1/e)^e), |a(n+1)/a(n)| = |ⁿ⁺¹x/ⁿx*hf(n,n)/hf(n+1,n+1)|, and ⁿx is again a two cycle, bounded above by 1 and below by 0. Using lemma #12, |a(n+1)/a(n)| < |ⁿ⁺¹x/ⁿx*1/ⁿ⁺¹n|, as before, and lim_m->+∞^2mx = a < 1, with a being a solution to x^xa=a, (See Solving the Second Real Auxiliary Exponential Equation), lim_m->+∞^2m+1x = b > 0, b < a, and with b given by x^a = b, above. This means that, |ⁿ⁺¹x/ⁿx|, n in N, is a two-cycle. We check the even and odd subsequences: lim_m->+∞|a(2m+1)/a(2m)| ≤ lim_m->+∞|^2m+1x/^2mx*1/^2m+1(2m)| = |b/a*0| = 0, while, lim_m->+∞|a(2m+2)/a(2m+1)| ≤ lim_m->+∞|^2m+2x/^2m+1x*1/^2m+2(2m+1)| = |a/b*0| = 0, and the lemma follows.

Notes

1. It is non-trivial to see what happens for x in (e^(1/e),+∞). Try to investigate yourself what the relation between k and x should be, so that lim_n->+∞|ⁿ⁺¹x/ⁿx*hf(n,k)/hf(n+1,k)| is 0, a > 0, or +∞.