Assume there is an off-line zeta zero
$$ \rho=\frac12+\delta+i\gamma, \qquad 0<\delta<\frac12. $$
Evaluate high logarithmic derivatives at the point directly to its right,
$$ s_0=1+\eta+i\gamma. $$
This is an old and very sensible idea. A high derivative turns a nearby pole into a large power. Turan's power-sum method then tries to isolate that large zero contribution and force an impossibly large prime sum.
The detector does amplify the zero exponentially.
It just cannot amplify it faster than the prime packet grows.
The exact arithmetic packet#
Write
$$ \xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s). $$
For every \(k\ge2\), the Hadamard product gives
$$ S_k(s) =\sum_\rho(s-\rho)^{-k} =\frac{(-1)^{k-1}}{(k-1)!} \left(\frac{\xi'}{\xi}\right)^{(k-1)}(s). $$
In \(\Re s>1\), this becomes the exact prime-side identity
$$ \boxed{ S_k(s) =s^{-k}+(s-1)^{-k} -2^{-k}\zeta(k,s/2)-P_k(s), } $$
where the third term uses the Hurwitz zeta function and
$$ \boxed{ P_k(s) =\frac1{(k-1)!} \sum_{n\ge2} \frac{\Lambda(n)(\log n)^{k-1}}{n^s}. } $$
So the arithmetic image of the zero power sum is not an abstract exponential polynomial. It is a very specific von Mangoldt packet.
Give the detector everything it wants#
Put
$$ c=\eta+\frac12, \qquad d=c-\delta. $$
The selected zero contributes exactly \(d^{-k}\) at \(s_0\). A critical-line zero at the same height would be horizontally \(c\) away.
Now be maximally generous. Assume a perfect Turan detector with no cluster loss, no gamma-factor loss, and no cancellation loss:
$$ \boxed{ |P_k(s_0)|\gtrsim d^{-k}. } $$
The coefficient-square norm of \(P_k\) has logarithmic saddle
$$ \log n=\frac{k}{c}+O(1), $$
so its square-root scale is
$$ \mathcal D_k(1+\eta)^{1/2} =c^{-k}\exp(o(k)). $$
The ideal squared gain over the diagonal mean-square scale is therefore
$$ \boxed{ \mathcal G_k =\left(\frac cd\right)^{2k}\exp(o(k)). } $$
This is real exponential amplification. The method is not failing because the pole signal is weak.
The absolute tail is longer#
To use a standard Dirichlet-polynomial mean-value theorem, truncate \(P_k\). A zero-independent absolute tail estimate has envelope
$$ y^{k-1}e^{-\eta y}\,dy, \qquad y=\log n, $$
whose saddle is
$$ y=\frac{k}{\eta}+O(1). $$
Thus the effective polynomial length is at least
$$ \boxed{ X_k=\exp(k/\eta+o(k)). } $$
Here is the entire obstruction:
$$ \boxed{ 2\log\frac cd < 2\log\frac c\eta =2\log\left(1+\frac1{2\eta}\right) < \frac1\eta. } $$
The detector's gain exponent is always smaller than the length exponent. Always. Every \(0<\delta<1/2\), every \(\eta>0\), every derivative order.
Higher moments do not rescue it#
A \(2q\)-th moment raises the optimistic signal to \(\mathcal G_k^q\), but also raises the length cost to \(X_k^q\).
If \(X_k^q\le T\), then \(qk\le\eta\log T+o(\log T)\), and
$$ \mathcal G_k^q \le T^{2\eta\log(c/d)+o(1)} =o(T). $$
The signal cannot outrun the height term.
If \(X_k^q>T\), then
$$ \frac{\mathcal G_k^q}{X_k^q} = \exp\left( qk\left[ 2\log\frac cd-\frac1\eta+o(1) \right]\right) \longrightarrow0. $$
The signal cannot outrun the length term either.
Let the cutoff chase the zero#
The absolute-tail estimate above is deliberately blunt. Perhaps the cutoff should be placed using the selected zero itself, retaining only the part of the packet that actually carries its signal.
For the sharply truncated packet
$$ P_{k,X}(s) =\frac1{\Gamma(k)} \sum_{n\le X} \frac{\Lambda(n)(\log n)^{k-1}}{n^s}, $$
the selected zero contributes exactly
$$ \boxed{ -d^{-k} \frac{\gamma(k,d\log X)}{\Gamma(k)}, } $$
where \(\gamma(k,\cdot)\) is the lower incomplete gamma function. Put
$$ \log X=\frac{rk}{d}. $$
The cutoff has now been optimized over every exponential scale, including choices that keep an exponentially tiny fraction of the full zero signal. If
$$ I_-(u)= \begin{cases} u-1-\log u,&0<u<1,\\ 0,&u\ge1, \end{cases} $$
then the selected-zero squared gain over the literal truncated diagonal has exponent
$$ \boxed{ E(r) =2\log\frac cd -2I_-(r) +2I_-\left(\frac{cr}{d}\right). } $$
The rude part survives optimization:
$$ \boxed{ E(r)<\frac rd \qquad(r>0). } $$
Before the diagonal saddle, the formula simplifies all the way to
$$ \boxed{ \mathcal G_k(r)=X^{2\delta+o(1)}=o(X), } $$
because \(2\delta<1\). After the saddle, the same strict exponent gap persists. So even a zero-adapted cutoff that is allowed to throw away almost all of the signal cannot make one selected zero beat the packet's ordinary length charge.
I find this wonderfully rude. Turan finds the hypothetical zero. It turns that zero into a genuinely huge prime-packet value. Then the packet needed to represent the value becomes even huger, by a strict exponent gap that no finite moment order can repair.
The verifier replays the completed power-sum identity through derivative order \(8\) with maximum error below \(2.3\times10^{-72}\), checks the exact single-pole square integral, checks the lower-gamma finite sum, and verifies the piecewise strict exponent gap across the full cutoff range.
What survives is sharply named: oscillatory shortening that uses the phase \(n^{-i\gamma}\), a theorem propagating one zero-induced large value to many ordinates, or a genuinely non-moment closure. What is closed is the standard and seductive chain
$$ \text{one zero} \longrightarrow \text{high derivative} \longrightarrow \text{one huge prime packet} \longrightarrow \text{finite-moment contradiction}. $$
The last arrow cannot pay its polynomial-length bill.
Notebook references: O-0309, O-0310