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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?

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4  
Did you try the usual functional equation (relating ζ(s) and ζ(1−s))? –  Carl May 7 '13 at 10:59
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Don't delete your question as I gave an answer. –  GH from MO May 7 '13 at 11:54
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up vote 10 down vote accepted

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.

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Thank you GH ${}{}$ –  joro May 7 '13 at 12:07
    
GH, can you express $\zeta''(-2)$ in simpler terms? Might have solution with zeta'(3),zeta(3),log,pi,gamma –  joro May 7 '13 at 13:52
    
@joro: It is possible to generate such expressions by looking at the Taylor series of $Z(s)=\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\zeta(s)$ at $s=-2$ and at $s=3$. The Taylor coefficients will be the same up to $\pm$ signs due to the symmetry $Z(s)=Z(1-s)$. Each Taylor coefficient in these identities can be expressed from the Laurent series coefficients at $s=-2$ and at $s=3$ of the factors involved in $Z(s)$. –  GH from MO May 7 '13 at 17:05
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