Is $\zeta^{(k)}(s)$, $k\geq 0$, negative for $s\in (0,1)$? It would seem to be the case that every derivative $\zeta^{(k)}(s)$ of the Riemann zeta function $\zeta(s)$ is negative for $s\in (0,1)$. (This is certainly the case for $k=0$, i.e., the Riemann zeta function itself: see How would you show that the Riemann Zeta function, $\zeta(s) < 0$ for $s \in (0,1)$?) What would be a quick way to show it? (Or is this at least in a standard reference?)
 A: Let's see. A good starting point is to have a series/integral representation for $\zeta(s)$ over the given region. For any $s>1$ we have
$$ \zeta(s)=\sum_{n\geq 1}\frac{1}{n^s} = \left(1-\frac{2}{2^s}\right)^{-1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} $$
and the domain of conditional convergence of the last series is larger, given by $\text{Re}(s)>0$.
It follows that for any $s\in(0,1)$ we may state
$$ \zeta(s) = \left(1-\frac{2}{2^s}\right)^{-1}\sum_{n\geq 1}(-1)^{n+1}\int_{0}^{+\infty}\frac{t^{s-1}}{\Gamma(s)}e^{-nt}\,dt $$
or
$$ \zeta(s) = \frac{1}{\Gamma(s)}\left(1-\frac{2}{2^s}\right)^{-1}\int_{0}^{+\infty}\frac{t^{s-1}}{e^t+1}\,dt $$
or (by integration by parts)
$$\boxed{ \zeta(s) = \frac{2^{2s-1}}{\Gamma(s+1)(2^s-2)}\int_{0}^{+\infty}\frac{t^{s}}{\cosh^2 t}\,dt }\tag{A}$$
Now comes the magic. $\int_{0}^{+\infty}\frac{t^{s}}{\cosh^2 t}\,dt$ is positive and log-convex by the Cauchy-Schwarz inequality, since it is a moment. $\frac{2^{2s-1}}{\Gamma(s+1)(2^s-2)}$ is negative, but its opposite $\frac{2^{2s-1}}{\Gamma(s+1)(2-2^s)}$ is log-convex too, by the Bohr-Mollerup theorem. So far the claim is proved up to $k=2$, and it is not difficult to extend such result. The key point is that all the coefficients of the Taylor series of $\log(-\zeta(s))$ at the origin, except the very first one, are positive. This can be proved by applying the very same approach of sliding-the-domain-of-conditional-convergence-by-one-to-the-left, by starting with
$$ \forall s:\text{Re}(s)>1,\qquad \log\zeta(s)=\sum_{n\geq 2}\frac{\Lambda(n)}{n^s\log(n)}.$$
But the most efficient approach is probably just to consider the logarithm of the Weierstrass product for $-\zeta(s)$:
$$\large\scriptstyle\log(-\zeta(s)) = -\log(2)-\log(1-s)+\frac{s}{2}\log(\pi e^{\gamma})+\sum_{n\geq 1}\left[\log\left(1+\frac{s}{2n}\right)-\frac{s}{2n}\right]+\sum_{\rho}\log\left(1-\frac{s}{\rho}\right)$$
The $-\log(1-s)$ term, associated with the simple pole of the $\zeta$ function at $s=1$, does most of the job. The other terms of the expansion provide, on occasion, negative contributions for the Taylor coefficients of the LHS, but such negative contributions are small, so the LHS essentially behaves like $-\log(2)-\log(1-s)$, and the claim is proved.
