zeta(2k+1) is a rational multiple of pi^{2k} zeta'(-2 k) ? 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?
 A: I am also sure that this is all known, but here is a quick proof that

$$ \zeta(2k+1)=\frac{(-1)^k2^{2k+1}}{(2k)!}\pi^{2k}\zeta'(-2k). $$

Following Carl Dettmann's suggestion, let us start from the functional equation
$$ \pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\zeta(s)=\pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)\zeta(1-s). $$
For $s=-2k+\epsilon$ this gives
$$ \zeta(2k+1-\epsilon)=\pi^{2k+\frac{1}{2}-\epsilon}\frac{\Gamma(-k+\epsilon/2)}{\Gamma(k+1/2-\epsilon/2)}\zeta(-2k+\epsilon). $$
On the right hand side, for $\epsilon\to 0$,
$$ \Gamma(-k+\epsilon/2)\sim \frac{2(-1)^k}{k!\epsilon}$$
and
$$ \zeta(-2k+\epsilon)\sim\epsilon\zeta'(-2k), $$
hence we have at $\epsilon=0$ i.e. at $s=-2k$
$$ \zeta(2k+1)=\pi^{2k+\frac{1}{2}}\frac{2(-1)^k}{k!\Gamma(k+1/2)}\zeta'(-2k). $$
On the right hand side
$$ \Gamma\left(k+\frac{1}{2}\right)=\left(k-\frac{1}{2}\right)\left(k-\frac{3}{2}\right)\cdots\frac{1}{2}\Gamma\left(\frac{1}{2}\right)=\frac{(2k-1)(2k-3)\cdots 1}{2^k}\pi^{\frac{1}{2}}, $$
whence the stated identity follows.
