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

[JmsNxn]

In MO the user Caleb Briggs brings to discussion an old attempt of mine to the series \( T(x,2) = 1^{1^x} - 2^{2^x} + 3^{3^x} - \cdots + \cdots \) .
I've tried this on 2007, and only later learned some techniques with which I might have assessed this with more success. But my basic observations and also the computations that I'd been able to do (over a small range of the exponent \(x\) ) come out to be correct.      

For the friends of visual data - there are some nice pictures to see there.

Here is the link : 

https://mathoverflow.net/questions/43666...eta-series   

Here the link to my fiddlings:

http://go.helms-net.de/math/tetdocs/Tetra_Etaseries.pdf  

Have fun...

Gottfried

@James :

I meant splitting the positive and negative parts of  \( T(x,2) = 1^{1^x} - 2^{2^x} + 3^{3^x} - \cdots + \cdots \)  and then using

 https://math.eretrandre.org/tetrationfor...p?tid=1688

my summability method on it.



regards

tommy1729

OHHHHH!

That makes much more sense Tommy  

So you're saying, let's continuum sum:

\[
E(z,s) = \sum_{j=1}^z (2j)^{(2j)^{-s}}\\
\]

And:

\[
P(z,s) = \sum_{j=1}^z (2j-1)^{(2j-1)^{-s}}\\
\]

Where then; Gottfried's function equals:

\[
\zeta_G(s) = \lim_{z\to\infty} \left(P(z,s) - E(z,s)\right)\\
\]

I mean, I don't see any reason this shouldn't work. Any continuum sum method works. But I'm not sure if this allows us to take \(-1 < \Re(s) \le 0\). Especially considering there is a pole at \(s=0\)--which should muddy the waters incredibly. I think the key though, may be somewhere in here. But we want to first find that value \(C \in \mathbb{R}\), such that:

\[
\zeta_G(s) - \frac{C}{s} = H(s)\\
\]

Where \(H(s)\) is holomorphic at \(s=0\). Because I don't think your summability method works in neighborhoods of poles (it's equivalent to the fractional calculus indefinite sum). And when there's a pole in the indefinite sum, it means an extra residue is added in the contour integral representation. Which muddies the waters so to speak.

But for \(\Re(s) > 0\)--your expression should work perfectly fine. And works for any indefinite sum. The function:

\[
f(z,s) = z^{z^{-s}}\\
\]

Is in the Ramanujan space \(f(z,s) = O(e^{\rho|\Re(z)| + \tau|\Im(z)|})\) for \(\rho, \tau \in \mathbb{R}^+\), and \(|\tau| < \pi/2\). But only for \(\Re(s) > 0\). I think this will suffer the same problem as my approach for \(-1 < \Re(s) < 0\). And I'm not sure yet how to handle this.

But:

\[
\sum_{j=1}^z f(j,s)\\
\]

Is perfectly indefinitely summable.