# Gottfried Helms' tetra-eta series

Here Gottfried Helms introduces the following fascinating divergent series $$T_2(x)=- \sum_{n=1}^\infty (-1)^n n^{n^x}$$ The terms don't go to zero, so technically the series does not converge anywhere. However, for $$\mathfrak{Re}(x)<0$$, the series is only diverging at a rate like $$n$$, so various weak divergent series methods can be applied (for instance, Cesaro or Abel summation) to get an answer. Applying one of these methods gives something that appears to have a dense set of zeroes near $$\mathfrak{Re}(x)=0$$, which can be seen in the following image (which has been rotated so that $$\mathfrak{Re}(x)=0$$ is the horizontal line at the top). This implies that $$\frac{1}{T_2(x)}$$ has a dense set of singularities on this line, thus $$T_2(x)$$ must also not be analytic at $$\mathfrak{Re}(x)=0$$. $$T_2(x)$$ has a very intricate and interesting structure, so I imagine there are many profound questions one can ask about this function. The particular question I am interested in is What does $$T_2(x)$$ look like when $$\mathfrak{Re}(x)\geq 0$$?

## $$T_2(x)$$ on the line $$\mathfrak{Im}(x)=0$$

The series diverges pretty quickly for $$0, and extremely quickly beyond there. Thus none of the usual divergent series summation methods work here. Nonetheless, Gottfried uses a novel approach with Euler summation to generate the following approximations for the sum We can extend where the series can be summed in the following way. Note that by the Residue theorem, we formally have that $$-\frac{1}{2i}\int_C \csc(\pi z) z^{z^x} dz = -\sum_{n=1}^\infty (-1)^n n^{n^x}$$. But we obviously cannot close the contour, since the $$z^{z^x}$$ will be very large for large $$z$$. However, for real values of x, $$z^{z^x}$$ is also becomes small in some directions. For instance, the following is a graph of $$z^{z^x}$$ near $$x=8$$. In this image, areas that are bright correspond to where the function is large, and dark areas are where it is small. In general, if we integrate along the contour that just covers only the first cone where the function is large (i.e. it doesn't go around any of the other cones), then we obtain exactly the same values Gotffried approximations suggest. For instance, the contour drawn in grey in the image above is an example of such a contour. Following this approach, we can extend Gottfried's table.

$$x$$ $$T_2(x)$$
$$\frac{1}{5}$$ $$0.252096$$
$$\frac{2}{5}$$ $$0.260102$$
$$\frac{3}{5}$$ $$0.271465$$
$$\frac{4}{5}$$ $$0.283753$$
$$\frac{5}{5}$$ $$0.295830$$
$$\frac{6}{5}$$ $$0.307391$$
$$\frac{7}{5}$$ $$0.318411$$
$$2$$ $$0.348736$$
$$3$$ $$0.392757$$
$$5$$ $$0.462847$$
$$10$$ $$0.563111$$
$$\infty$$ $$1$$

We can confirm the value at $$-\infty$$, $$-1$$, $$0$$, $$1$$ and $$\infty$$ are correct. At $$-\infty$$ the sum is equal to $$-\sum_{n=1}^\infty (-1)^n = \eta(0)=\frac{1}{2}$$. At $$-1$$, the sum is equal to $$\sum_{n=1}^\infty (-1)^n n^{\frac{1}{n}}$$, and is equal to the average of the two MRB constants (see here). At $$0$$, the sum is $$-\sum_{n=1}^\infty (-1)^n n = \eta(-1) = \frac{1}{4}$$. At $$x=1$$ the constant is $$1-\int_1^\infty \frac{1}{x^x} dx$$ which is discussed here. At $$x=\infty$$ only the first term is not infinite, and usually with divergent series this means the sum must be equal to the first term, which is $$1^{1^\infty} = 1$$.

A graph of the function looks like this: Unfortunately, my approach doesn't work when $$x$$ has an imaginary part. Thus, I am interested in any approaches that can assign values to complex values of $$x$$. I am also curious about the behavior of this function at $$\mathfrak{Re}(x)=0$$. For instance, does the sum converge to some distribution? Also-- are there any functional or algebraic equations that $$T_2(x)$$ satisfies, and is there perhaps some relationship between $$T_2(x)$$ and $$T_2(-x)$$?

## $$T_2$$ on the line $$\mathfrak{Re}(x)=0$$

It turns out that it is possible to sum the series on this line as well, using approximately the same approach. Here the contour we take goes from $$c+\infty i$$ to $$c$$, and then and then $$c + \infty e^{-\varepsilon i}$$, with $$0 and the imaginary part is positive. Doing this gives the following parametric plot for $$z = it$$ with $$-6. Interestingly, it doesn't seem to have any zeroes, at least checking up to $$|t|=15$$. Thus, a possible conjecture is that the zeroes of the function lie between the strips $$\mathfrak{Re}(z)=-2$$ and $$\mathfrak{Re}(z)=2$$.

### Update

Since $$\frac{n^{n^x}}{n!} \to 0$$ when $$\mathfrak{Re}(x)<1$$, the series is Borel summable in this area (but not beyond here). Doing some calculations, it appears that $$T_2(x)$$ doesn't actually have a natural boundary at $$\mathfrak{Re}(x)=0$$, but rather at $$\mathfrak{Re}(x)=1$$. Although the location of the earlier zeroes appears correct, it doesn't seem that they accumulate densely on the line $$\mathfrak{Re}(x)=0$$ as I increase the number of terms in the approximation as I had initially suspected. I've put together a graph that computes the function up $$\mathfrak{Re}(x)=1$$, though my current computation doesn't have enough precision and starts to break down near $$\mathfrak{Re}(x)=1$$. ### Update 2

I recently noticed that $$\sum_{n=1}^\infty (-1)^n n^{n^s} = \sum_{k=0}^\infty \frac{1}{k!} \sum_{n=1}^\infty \ln(n)^k n^{sk} = \sum_{k=0}^\infty \frac{(-1)^k}{k!} \eta^{(k)}(-sk)$$ This function agrees with the other calculations on this question. It also allows us to take a step towards computing the much more difficult series $$\sum_{n=1}^\infty n^{n^s} = \sum_{k=0}^\infty \frac{(-1)^k}{k!} \zeta^{(k)}(-sk) + \mathbf{??}$$ where the missing question mark term should come from residue of the zeta function (which isn't present in the eta function). Computing exactly what the $$\mathbf{??}$$ function is seems difficult, however, since it seems to require extending the zeta function with complex-order derivatives.

• Concerning $z^{z^x}$ - it might be more manageable to study its double logarithm: it is equal to $xe^t+t$, where $t=\log\log z$. Dec 16, 2022 at 6:32
• In fact just a single logarithm already seems reachable: it is $ue^{xu}$, where $u=\log z$. Dec 16, 2022 at 6:37
• Hi Caleb - very sweet to see that paper been discussed! I've had this nearly forgotten because of the difficulties in the summation; now you show a much extended range for $x$ - very cool. The pictures are great - - I'm going to chew on this. - Just a little bug (?): in the "update 2" formulae we have on the lhs an "x" which is missing in the rhs, and an "s" in the rhs which is missing in the lhs. Perhaps a typo? Jan 28 at 11:57
• @GottfriedHelms It is truly a fascinating function-- it has been in the back of my mind for a little over a year and I was excited when I recently developed techniques to sum it. Also, you are correct, there is a typo in update 2 that I will fix Jan 29 at 1:44
• Caleb - unfortunately I feel no more able to step inside this with concepts which are new to me (at least indepth aspects, of course I know something freshman's exercises about integration). But perhaps what I can do would be to introduce you to complex height tetration using Schroeder's/Koenig's method if applicable to the function at hand. For beyond-introduction-questions some members in the "tetration-forum" are usually kind advisors... Jan 29 at 12:19