Discussion on "tetra-eta-series" (2007) in MO

[Caleb]

AHHH! I see, I apologize. I was confused!

Then yes we are on the exact same page  

I've made some graphs I'll post in a bit, but the zeta series appears to converge for \(0 < \Re(s) \), where I presume we can analytically continue using Gottfried's expansion. I apologize!

If you are saying that:

\[
\sum (-1)^n n^{n^{-s}}
\]

converges for \(\Re(s) > -1\) we are on the same page. I actually haven't looked at this sum, I just instantly wrote it as a zeta function and only looked at that--so I wasn't sure where the original series converged.

It also appears the zeroes are on \(\Re(s) = 1/2\) on the critical strip... having Riemann Hypothesis vietnam flashbacks


So I accidentally closed one of the graphs mid compile, but the other one finished. So here is:

\[
\nu(s) = \sum_{m=1}^\infty q(m) m^{-s}\\
\]

Where:

\[
q(m) = (-1)^m \sum_{n \le m} \chi_m(n)\\
\]

And \(\chi_m(n) =1\) if and only if \(n^k = m\) for some \(k \in \mathbb{N}\).

Then here is a graph from:

\(0.5 \le \Re(s) \le 5.5\) and \(|\Im(s)| \le 2.5\):



Didn't finish the graph for gottfried's function, it only finished most of the top half, so I'm going to recompile. I am also graphing the critical strip for both, to see what that could look like.

Good to see everything is set straight now. Also, to ease communication, I'll refer only to \( \sum (-1)^n n^{n^{-s}} \) so we are looking at the same object.

To make sure I understand your methods, initially 
\[ P(s) = \sum g(m) m^{-s} \]
converges only for \(\Re(s)<0\), otherwise the terms don't go to zero. However, using some complex analysis (abel sum formula), we can write 
\[ P(s) = -s \int_1^\infty A(x)x^{-s-1}\]
Then if we assume \( A(x) \) has growth smaller than a polynomial (in particular, it looks like \( A(x) \) has growth like the logarithm), then the second formula analytically continues the sum up to the point where \( \Re(s) > -1 \). 

Another thing to note, is that as \( \Re(s) \to -1 \) we get that the integral takes longer and longer to converge, so theoretically it should take on more and more of the choas of \( A(x) \) [In particular, if \( A(x) \) were completely random, then we would be forced to have a natural boundary, since nearby values of s would lead to very different results, so \( A(x) \) would be almost completely random]. If I understand correctly, questions about whether the function has a natural boundary should reduce into how 'predictable' \( A(x) \) is.