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Is it true that $(-1)^{n-1} {(1/\zeta)}^{(n)}(x) >0$ for all real $x>1$ ? Or in other words can you show that the higher order derivatives of the reciprocal of the Riemann zeta function alternate in sign when $x>1$?

This obviously true for the Riemann zeta function but I couldn't prove it for the reciprocal since mobius function changes its sign and does not guarantee positivity or negativity of the expression.

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In other words, you ask whether the function $f(x):=1-1/\zeta(1+x)$ is completely monotonic on $[0,+\infty)$. We have $f(x)=\sum_{n>1} -\mu(n)/n^{1+x}=\int e^{-xt}d\lambda(t)$, where $\lambda=\sum_{n>1} -\frac{\mu(n)}n\delta_{\log n}$. Alas, this measure is not positive (say, $\lambda(\{6\})=-1/6$), and by uniqueness of inverse Laplace transform and Bernstein theorem your assumption does not hold.

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    $\begingroup$ Do you mean $\lambda(\{\log 6\})$? What looks a little strange to me, your argument only uses $\mu(6)>0$, no other coefficient of the Dirichlet series $1/\zeta(x)$ matters? $\endgroup$
    – Peter Mueller
    Commented Nov 23 at 16:21
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    $\begingroup$ @PeterMueller Yes, one negative coefficient is enough to this assumption to fail. Because for large $n$ and $x=n/\log 6$ the $n$-th derivative of $6^{-x}$ is the main one. $\endgroup$
    – Fedor Petrov
    Commented Nov 23 at 16:32
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    $\begingroup$ Ah, I see. Using $\mu(k)\ge-1$ for $k\ge7$ in the series of $1/\zeta(x)$, one obtains $(-1)^n(1/\zeta(x))^{(n)}\ge\log(4)^n/4^n+2\log(6)^n/6^n-(-1)^n\zeta(x)^{(n)}$. And from that one obtains that the right hand side is positive for $x=n/\log(6)$ if $n$ is big enough (like $n=188$). Very nice indeed! $\endgroup$
    – Peter Mueller
    Commented Nov 23 at 17:40
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    $\begingroup$ @PeterMueller Yes I spent around a month thinking that it is true in general but I didn't expect it to fail. I might ask another question about my other conjecture since I don't want to waste time thinking that it is true again :). $\endgroup$
    – Haidara
    Commented Nov 23 at 18:01
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    $\begingroup$ @PeterMueller Also thank you for your counter example. $\endgroup$
    – Haidara
    Commented Nov 23 at 18:01
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Here is a more pedestrian version of Fedor Petrov's argument: We set $\lambda=1/\log6$ and claim that $(-1)^n(1/\zeta)^{(n)}(\lambda n)>0$ if $n$ is big enough. With $a_k=\log k/k^\lambda$, we have \begin{equation} (-1)^n(1/\zeta)^{(n)}(\lambda n) =\sum_{k\ge1}\mu(k)a_k^n\ge a_6^n-\sum_{k\ne6}a_k^n, \end{equation} where we used $\mu(6)=1$ and $\mu(k)\ge-1$.

One quickly checks that $0\le a_k/a_6<1$ for $k\ne 6$ (because $\log x/x^\lambda$ assumes a strict maximum in $x=6$.) So the claim follows once we know that $\sum_{k\ne6}(a_k/a_6)^n<1$ for all $n$ big enough. Set $c=1/\zeta(2)$. For $n$ big enough, we have $(a_k/a_6)^n<c/k^2$ for all $k\ne6$, because this holds for $k=1$ (recall that $a_1=0$) and for $2\le k\ne6$ this is equivalent to \begin{equation} n>\frac{2\log k-\log c}{\log a_6-\log a_k} =\frac{2\log k-\log c}{\log a_6+\lambda\log k-\log\log k} \end{equation} and the right hand side is bounded (it converges to $2/\log\lambda$).

We get $\sum_{k\ne6}(a_k/a_6)^n\le\sum_{k\ne6}c/k^2<c\zeta(2)=1$, and the claim follows.

Remark: With these rough inequalities, the OP's conjecture fails for all $n\ge 970$. One can tweak the proof at various places. But it seems to me that the smallest $n$ where the conjecture fails is around $n=155$.

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    $\begingroup$ Thank you. I appreciate your work. $\endgroup$
    – Haidara
    Commented Nov 24 at 12:48

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