A Strange Hyperexponential Series

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

In this article, let's try to investigate the tetration series:

n=11/nx = 1/1x + 1/ 2x + 1/ 3x + ..., for real x. (1)

The hyperexponentials on the denominators grow very fast, so we expect it to converge for some values of x. Our basic tools are the two articles on convergence of hyperexponentials, Infinite Exponentials and A Deeper Analysis of Infinite Exponentials. Knowing that limn->+∞nx exists from the first article for x in [(1/e)e,e(1/e)], we expect immediately the series to have problems there. Let us see what happens.

Lemma #1:

The hyperexponential series diverges for x in [(1/e)e,e(1/e)].

Proof:

For x in the indicated interval, the n-th term of the series is 1/nx, and the limit of this is (see article 1) limn->+∞1/nx = 1/limn->+∞nx = 1/e-W(-log(x))= eW(-log(x)), which does not approach 0, therefore the series violates the Cauchy criterion. Accordingly, it diverges there.

Lemma #2:

The hyperexponential series converges for x > e1/e.

Proof:

In this case, we apply the Ratio Test. First note that x > e1/e => Log(x) = A > 1/e > 0. (1)

limn->+∞[an+1/an] = limn->+∞[nx/n+1x] = limn->+∞[nx/x(nx)] = limt->+∞t/xt =  limt->+∞t/eLog(x)*t = limt->+∞t/eA*t = 0 < 1, (A > 0, by (1)), since the terms nx = t are unbounded, so the series converges there.

Lemma #3

The hyperexponential series diverges for 0 < x < (1/e)e.

Proof:

For x on the indicated interval, the infinitely iterated exponential limn->+∞nx is a two-cycle. (See article 1). But the partial sums of the series (1) are bounded below by the series with even terms, i.e. ∑n=1[t/2]1/2nx < ∑n=1t1/nx, and the first series diverges, because it again violates the Cauchy criterion, since limn->+∞2nx = a, with a a solution to: xxa=a (see Solving the Second Real Auxiliary Equation), therefore limn->+∞1/2nx = 1/a =/= 0, and the Lemma is proved.

Observations:

Setting up exact Maple code to evaluate this series when x is in (e1/e, +∞) is almost impossible, since if x is away from e1/e, the denominators grow so large that Maple is unable to handle the quotients. However we can set up code that approximates the series pretty well. First, we can modify the function f_N to handle quotients a bit better. Refer to the code for f_N on this article on hyperroots.

> f_NR:=proc(z,w,n)
> option remember;
> if n=1 then 1/z^w;
> else 1/z^(1/f_NR(z,w,n-1));
> fi;
> end:

And the series:

> f:=x->sum('f_NR(x,1,n)',n=1..5);

We can vary the final bound of the n above, depending on how large our values of x are. With n=5, we can perhaps calculate:

> evalf(f(2));
.8125152588

But anything larger will fail. Try:
> evalf(f(2.1));
(integer too large)

If however we lower the bound for n still more, down to 3, we can see some of the values up to 2.8.

> evalf(f_NR(2,1,4));
.00001525878906

While for x > 3 the terms of the series approach 0 very fast. In fact:

> evalf(f_NR(3,1,3));
.1311372652 10-12

We can therefore approximate the series pretty well using the following code:

> f:=proc(x)
> if x >= 2 then
>  sum('f_NR(x,1,n)',n=1..3);
> elif(1.5<x) and (x<2) then
>  sum('f_NR(x,1,n)',n=1..5);
> elif (evalf(exp(exp(-1)))<x) and (x<1.5) then
>  sum('f_NR(x,1,n)',n=1..10);
> else print(`series diverges`);
> fi;
> end:

> with (plots):
> plot('f(x)',x=evalf(exp(exp(-1)))..5);

hyperexponential series graph

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