[NT] more zeta stuff for the fans

[JmsNxn]

On mathstack the following conjecture about a dirichlet series was written.

https://math.stackexchange.com/questions...dot3-s-5-c

Since there are " zeta fans " here ...


Regards

tommy1729

I don't think this is extendable beyond \(\Re(s) > 2\)--and if it were it would be very non-obvious. To ask for zeroes before analytically continuing is a recipe for disaster. The sum DOES NOT converge for \(1 < \Re(s) < 2\)--so to say there are zeroes here; when the sum diverges; without an analytic continuation is... nonsense.

The standard way to analytically continue is useless here too. If I write:

\[
P(x) = \sum_{n\le x} p_n\\
\]

Then the tight bounds are:

\[
P(x) = \frac{x^2}{2} \left(\log(x) + \log \log(x) - 3/2 + o(1)\right)\\
\]

So from here we have:

\[
\zeta_P(s) = -s\int_1^\infty P(x)x^{-s-1}\,ds\\
\]

This integral converges for \(\Re(s) > 2\); but once you account for \(P(x)\)'s asymptotics--there's no obvious way to extend it. The error term \(o(1)\) decays too slowly-- about \(1/log^2 x\) to meaningfully allow for a simple continuation. Unless mick has a magical reflection formula up his sleeve, this is pretty pointless...

But that integral transform is designed not to have analytic continuation.

If P(x) = x^3 this would also not converge , yet its related zeta function does as it can be expressed by riemann zeta functions.



In fact if we write p_n as a taylor series 

p(n) = a_0 + a_1 n + a_2 n^2 + ...

Then we get expressions like

A(s) = zeta(s) + a_1 zeta(s-1) + a_2 zeta(s-2) + a_3 zeta(s-3) + ...

which might have analytic continuation then, since zeta does.


Notice this is not like an euler product or the prime zeta function.



And even if we have an natural boundary , that does not explain why the zero's approach the boundaries.


regards

tommy1729

I mean, I kind of see where your coming from.

But if \(|A_M(s) - A(s)| < \epsilon/2\) and \(|A(s)| > \epsilon\)--then:

\[
\begin{align}
|A(s)| &\le |A_M(s)| + |A_M(s) - A(s)|\\
|A_M|(s) &\ge \epsilon/2\\
\end{align}
\]

Where \(\epsilon\) only depends on \(M\) and \(\Re(s)>2\).

---

This is exactly what they teach you NOT TO DO! to solve the Riemann Hypothesis. If I take the sum:

\[
\zeta^M(s) = \sum_{n=1}^M n^{-s}\\
\]

The zeroes of this function coallesce near \(\Re(s) = 1\); because the sum goes to infinity and back near here. But it has no marker on the actual zeroes of \(\zeta(s)\). Because \(\zeta\) NEEDS TO BE analytically continued first... Without an analytic continuation; we can't speak on the zeroes. The partial sums tell us nothing about the zeroes; because the partial sums converge to infinity as \(M\to\infty\) in the critical strip--there is no domain of normality.