Probably this is well know and elementary and will delete it, but couldn't find it on the web.

Got a sketch of proof and numerical evidence that $\zeta(2k+1)$ is a rational multiple of $\pi^{2k} \zeta'(-2k)$

An identity from Derivatives of the Hurwitz Zeta Function for Rational Arguments p.7

$$ \zeta'(-n,x) + (-1)^n \zeta'(-n,1-x) = \pi i \frac{B_{n+1}(x)}{n+1} + \frac{n!}{(2 \pi) ^ n} e^{-\pi i n / 2} \operatorname{Li}_{n+1}(e^{2\pi i x}). \qquad (21)$$

For even $n$ and $x=\frac12$ (21) is:

$$ 2 \zeta'(-n,\frac12) = \pi i \frac{B_{n+1}(\frac12)}{n+1} \pm \frac{n!}{(2 \pi) ^ n} \operatorname{Li}_{n+1}(-1).$$

The choise of $\pm$ depends on $e^{-\pi i n / 2}$.

According to Wolfram Alpha $Li_{n+1}(-1)$ is an integer multiple of $\zeta(n+1)$ and $B_{n+1}(\frac12)$ vanishes.

$\zeta(s,\frac12) = (2^s-1) \zeta(s) $. Taking derivative and having in mind $\zeta(-2k)=0$ we have $\zeta'(s,\frac12)$ is a rational multiple of $\zeta'(s)$.

For even $n$ and $x=\frac12$ (21) simplifies to:

$$ \mathbb{Q} \zeta'(-n) = \pm \mathbb{Z} \frac{n!}{(2 \pi) ^ n} \zeta(n+1). \qquad (1)$$

In particular,

$$ \zeta(3) = -4 \zeta'(-2) \pi^2$$ $$ \zeta(5) = 4/3 \zeta'(-4) \pi^4$$

The last two hold with precision of 1000 digits.

Is this true?

Is this known?