Is the following inequality true?
$$\gamma -\frac{\zeta ''(-2\;n)}{2 \zeta '(-2\;n)} > \log (n)-\gamma$$
This for $n$ a positive integer, $n=1,2,3,4,5,…$, and more precisely when $n$ approaches infinity.
$\gamma$ is the Euler-Mascheroni constant, $\zeta(s)$ is the Riemann zeta function and $\zeta'(s)$ is a derivative.
Plotting the left hand side (red) and the right hand side (blue):
we see that they are close to each other. Mathematica program for the plot:
Clear[n, s, t, aa, bb, gg1, gg2, g1, g2, nn, a, b]
nn = 40;
Limit[Zeta[s] -
Derivative[1][Zeta][s - 1 + ZetaZero[1]]/Zeta[s - 1 + ZetaZero[1]],
s -> 1]
a = Monitor[
Re[Table[
EulerGamma -
Derivative[2][Zeta][-2*n]/(2*Derivative[1][Zeta][-2*n]), {n, 1,
nn}]], n];
Clear[n]
Print["Log[n]-EulerGamma"]
b = Table[N[Log[n]] - EulerGamma, {n, 1, nn}];
g1 = ListLinePlot[a, PlotStyle -> Red];
g2 = ListLinePlot[b];
Show[g1, g2, ImageSize -> Large]
Edit 21 4 2014: After Raymond Manzonis answer below:
$$\Im(\rho _n) \sim -\frac{1}{2} \left(-\frac{\zeta '(\Im(\rho _n)+1)}{\zeta \left(\Im\left(\rho _n\right)+1\right)}-\psi \left(\Im\left(\rho _n\right)+1\right)+\log (2 \pi )\right)+\frac{1}{\Im\left(\frac{1}{\rho _n}\right)}-\Re\left(\frac{\zeta ''(\rho _n)}{2 \zeta '(\rho _n)}\right)$$
nn = 40;
-N[Table[Re[(Zeta''[
ZetaZero[n]]/(2*Zeta'[ZetaZero[n]]))] - (((Log[2*Pi] -
PolyGamma[Im[ZetaZero[n]] + 1] -
Zeta'[Im[ZetaZero[n]] + 1]/Zeta[Im[ZetaZero[n]] + 1])/2)
) + (Im[ZetaZero[n]^-1]^-1), {n, 1, 12}], 30]
(*program start*)(*After Raymond Manzoni Mathematics stackexchange \
Apr 20 2014*)
nn = 30;
N[Table[-((Re[(Zeta''[
ZetaZero[n]]/(2*Zeta'[ZetaZero[n]]))] - (((Log[2*Pi] -
PolyGamma[Im[ZetaZero[n]] + 1] -
Zeta'[Im[ZetaZero[n]] + 1]/Zeta[Im[ZetaZero[n]] + 1])/
2)) + (Im[ZetaZero[n]^-1]^-1) + Im[ZetaZero[n]])*
Im[ZetaZero[n]]^2), {n, 1, nn}], 20]
%^-1
Clear[n, x, y, a]
a = x /. Solve[-((y + x)*x^2) == 1/48, x][[2]]
n = 62;
y = Re[(Zeta''[
ZetaZero[n]]/(2*Zeta'[ZetaZero[n]]))] - (((Log[2*Pi] -
PolyGamma[Im[ZetaZero[n]] + 1] -
Zeta'[Im[ZetaZero[n]] + 1]/Zeta[Im[ZetaZero[n]] + 1])/
2)) + (Im[ZetaZero[n]^-1]^-1);
(N[a, 30] -
Im[N[ZetaZero[n], 30]])*Im[N[ZetaZero[n], 30]]^4
Re[%]^-1
Just for memory May 3, 2014:
Clear[s, z]
Monitor[Table[
Limit[Zeta[s] -
RiemannSiegelZ'[s - 1 + Im[ZetaZero[z]]]/
RiemannSiegelZ[s - 1 + Im[ZetaZero[z]]], s -> 1], {z, 1, 4}], z]
N[%]
Table[N[Im[
1/2 I Log[\[Pi]] -
1/4 I PolyGamma[0, 1/4 - 1/2 I Im[ZetaZero[n]]] -
1/4 I PolyGamma[0, 1/4 + 1/2 I Im[ZetaZero[n]]]] ==
Im[(I Zeta''[1/2 + I Im[ZetaZero[n]]])/(
2 Zeta'[1/2 + I Im[ZetaZero[n]]])]], {n, 1, 12}]
Returns True,True,True,…
Best Answer
Trivial zeros
Numerically I found that for $n$ any positive integer (replacing your $\,\gamma+\gamma\,$ constant by $\,\log\,\pi\,$) : $$\tag{1}\frac{\zeta ''(-2\;n)}{2\,\zeta '(-2\;n)}+\log (n)<\log(\pi)$$
with the limit approaching $\log\,\pi\,$ as $\,n\to \infty$.
I obtained too the following asymptotic expansion as $\,n\to\infty$ : $$\frac{\zeta ''(-2\;n)}{2\,\zeta '(-2\;n)}+\log\left(\frac n{\pi}\right)\sim -\frac 1{4\,n}+\frac 1{48\,n^2}-\frac 1{1920\,n^4}+\frac 1{16128\,n^6}-\frac 1{61440\,n^8}+\operatorname{O}\left(\frac 1{n^{10}}\right)$$ that should probably be the nice :
$$\tag{2}\frac{\zeta ''(-2\;n)}{2\,\zeta '(-2\;n)}+\log\left(\frac n{\pi}\right)\sim \sum_{k=1}^\infty \frac {B_k}{k\,(2n)^k}$$ (this is an asymptotic expansion and $k$ shouldn't really go up to infinity)
At this point we may use this formula for the digamma function : $$\tag{3}\psi(z+1)\sim \log(z)-\sum_{k=1}^\infty \frac {B_k}{k\,z^k}$$ (the '$+1$' in $\,\psi(z+1)=\psi(z)+\dfrac 1z\,$ doesn't appear in Wikipedia because of their alternative convention $B_1:=\dfrac 12$ instead of $B_1:=-\dfrac 12$)
with $\,z:=2n\,$ to rewrite $(2)$ with high precision (error $< 10^{-60}$ for $n=100$) as : $$\tag{4} \frac{\zeta ''(-2\;n)}{2\,\zeta '(-2\;n)}\sim \log(2\pi)-\psi(2n+1)$$
All this may be proved starting with the logarithmic derivative of the functional equation of $\zeta$ (exchanging $s \leftrightarrow (1-s)$ first) : $$\left[\log\,\zeta(1-s)\right]'=\left[\log\left(2(2\pi)^{-s}\Gamma(s)\cos\left(\frac{\pi}2s\right)\zeta(s)\right)\right]'$$ that is, since $\,\psi(s):=[\log\,\Gamma(s)]'\,$ and putting $\,-\log\left(\cos\left(\frac{\pi}2s\right)\right)'$ at the left : $$\tag{5}\frac{\pi}2\tan\left(\frac{\pi}2 s\right)-\frac{\zeta'(1-s)}{\zeta(1-s)}=-\log(2\pi)+\psi(s)+\frac{\zeta'(s)}{\zeta(s)}$$ the left part becomes in the 'degenerate' case (i.e. $\zeta(1-s)\to 0$) $s:=2n+1$ : $$\lim_{h\to 0}\frac{\pi}2\tan\left(\frac{\pi}2 (2n+1+h)\right)-\frac{\zeta'(-2n-h)}{\zeta(-2n-h)}=\lim_{h\to 0}-\frac 1h+O(h)+\frac 1h-\frac{\zeta''(-2n)}{2\,\zeta'(-2n)}+O(h)$$ and minus $(5)$ gives the exact (for $n$ positive integer!) : $$\tag{6}\boxed{\displaystyle\frac{\zeta''(-2n)}{2\,\zeta'(-2n)}=\log(2\pi)-\psi(2n+1)-\frac{\zeta'(2n+1)}{\zeta(2n+1)}}$$ (note that $\;\psi(2n+1)=H_{2n}-\gamma\,$ with $H_m$ the $m$-th harmonic number)
But $(6)$ is exactly $(4)$ with an additional term $-\frac{\zeta'(2n+1)}{\zeta(2n+1)}$ that may easily be shown very small for $n\gg 1$ implying thus $(4)$ and the earlier equivalences, the claimed limit and the initial inequality.
Non trivial zeros, critical line
In this case the singularity of $\dfrac{\zeta'(1-s)}{\zeta(1-s)}$ is not canceled by the singularity of the tangent but by the singularity of the other fraction $\dfrac{\zeta'(s)}{\zeta(s)}$ (for $\,0<\Re(s)<1\,$ the functional equation implies the equivalence between $\,\zeta(s)=0\,$ and $\,\zeta(1-s)=0$).
The same method as for $(6)$ allows us to get for $\rho$ any nontrivial zero : $$\tag{7}\lim_{s\to \rho}\frac{\zeta'(1-s)}{\zeta(1-s)}+\frac{\zeta'(s)}{\zeta(s)}=\frac{\zeta''(1-\rho)}{2\,\zeta'(1-\rho)}+\frac{\zeta''(\rho)}{2\,\zeta'(\rho)}$$
In the usual case $\,\rho=\dfrac 12+i\,t\;$ with $t\in\mathbb{R}$ (all known non trivial zeros!) we have $\,1-\rho=\overline{\rho}\,$ allowing us to rewrite the right side of $(7)$ simply as $\,\displaystyle \Re\left(\frac{\zeta''(\rho)}{\zeta'(\rho)}\right)\,$ as the exact (for $\rho$ any nontrivial zero and $(5)$) :
$$\tag{8}\boxed{\displaystyle\Re\left(\frac{\zeta''(\rho)}{\zeta'(\rho)}\right)=\frac{\pi}2 \tan\left(\frac{\pi}2 \rho\right)+\log(2\pi)-\psi(\rho)}$$ while for $\,\rho=\dfrac 12+i\,t\;$ not a zero on the critical line (defined by $\,\Re(\rho)=\frac 12$) : $$\tag{9}\boxed{\displaystyle 2\;\Re\left(\frac{\zeta'(\rho)}{\zeta(\rho)}\right)=\frac{\pi}2 \tan\left(\frac{\pi}2 \rho\right)+\log(2\pi)-\psi(\rho)=\,-2\,\theta'(t)}$$ with $\theta(t)$ the Riemann-Siegel theta function since the logarithmic derivative of $\,\displaystyle\zeta\left(\frac 12+it\right)=Z(t)\,e^{-i\theta(t)}\;$ is $\quad\displaystyle i\frac {\zeta'\left(\frac 12+it\right)}{\zeta\left(\frac 12+it\right)}=\frac{Z'(t)}{Z(t)}-i\theta'(t)$.
For $\,t \gg 1\,$ and since the real part of the tangent term becomes quickly very small we have :
$$\tag{10}-2\,\theta'(t)\approx \log(2\pi)-\Re\,\psi\left(\frac 12+it\right)$$ The page $259$ of Abramowitz & Stegun contains the equation $6.3.8$ and gives $\,\displaystyle \psi\left(\frac 12+it\right)=2\psi(1+2it)-\psi(1+it)-\ln(4)\,$ while the asymptotic expansion $6.3.19$ will give a similar expression : $$\Re\,\psi\left(\frac 12+it\right)\sim\log(t)-\sum_{n=1}^\infty\left(1-\frac 1{2^{2n-1}}\right)\frac{(-1)^{n-1}B_{2n}}{2n\,t^{2n}}$$ and the fine approximation : $$\tag{11}-2\,\theta'(t)\approx \log\left(\frac{2\pi}t\right)+\frac 1{24\,t^2}+\frac 7{960\,t^4}$$ All this shows that the derivative of the argument of $\zeta$ will be very smooth on the critical line (as well as $\theta(t)$ itself after integration) while the left term of $(8)$ $\,\Re\frac{\zeta''(\rho)} {\zeta'(\rho)}\,$ will be smooth only near the nontrivial zeros (elsewhere $Z(t)$ and its derivatives will be involved!).
Let's illustrate this by showing the left (irregular) $\,\Re\frac{\zeta''(\rho)} {\zeta'(\rho)}$ part and right (regular) $-2\theta(t)$ part of $(8)$ for $1<t<50$ :
If we subtract these two curves we will obtain a new curve taking the value $0$ only when the formula $(8)$ applies. The second curve is the real Riemann Z-function (i.e. $\zeta$ 'without the phase') taking the value $0$ only at the imaginary parts of the zeros :
In the $(1,50)$ range considered $(8)$ has only one additional solution for $t$ a little smaller than $40$.