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    $\begingroup$ @Speyer: "I don't know why Mathematica thinks it's okay to plug in $s=0$." Secretly Mathematica is a physicist? $\endgroup$ Commented Mar 12, 2013 at 15:18
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    $\begingroup$ Well we expect the formula for $-\zeta'/\zeta^2$ to hold for ${\rm Re}(s) > 1/2$... But certainly not for $s=0$. $\endgroup$ Commented Mar 12, 2013 at 15:20
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    $\begingroup$ One way to regularize $\sum_{n=1}^N f(n)$ is (from terrytao.wordpress.com/2010/04/10) by taking a cutoff function $\eta$ with compact support satisfying $\eta(0) = 1$, and using $\sum_{n=1}^\infty \eta(n/N) f(n)$. For $\eta = \chi_{[0,1]}$, you get the ordinary sum, and for $\eta(x) = (1-x)\chi_{[0,1]}$ you get Cesáro. If you want a good asymptotic expansion, $\eta$ should be smooth, or else boundary effects crop up. At any rate, you get your value of $2 \log(2\pi)$ plus $\sum_\rho c_\rho N^\rho$ as $\rho$ varies over zeroes of $\zeta'$, which I don't know how to control. $\endgroup$ Commented Mar 13, 2013 at 2:14
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    $\begingroup$ @David Speyer, Cesaro summability cannot be true because in that case we would have an analytic continuation of $\zeta'/\zeta^2$ on $\textrm{Re}(s)>0$. $\endgroup$ Commented Mar 18, 2013 at 3:12
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    $\begingroup$ @Jason Starr Mathematica uses formal series, that is why this answer is ok for it. $\endgroup$ Commented Dec 14, 2013 at 2:47