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The following analytic continuation for $\zeta(s)$ towards $\Re(s)>-1$ was derived here:

$$\displaystyle \zeta(s) = \frac{1}{2\,(s-1)} \left(\sum _{n=1}^{\infty } {\frac {s-1-2\,n}{{n}^{s}}} + \sum _{n=0}^{\infty } {\frac {s+1+2\,n}{\left( n+1 \right) ^{s}}} \right)$$

I plugged this formula into this question about the zeros of $\zeta(s) \pm \zeta(1-s)$. After rearranging the terms a new function emerges:

$$\displaystyle Z(s) = \frac{1}{2\,(s-1)} \sum _{n=1}^{\infty } {\frac {s-1-2\,n}{{n}^{s}}} \pm \frac{1}{2\,(-s)} \sum _{n=0}^{\infty } {\frac {1-s+1+2\,n}{\left( n+1 \right) ^{1-s}}}$$

with $\zeta(s) \pm \zeta(1-s) = Z(s) \pm Z(1-s)$.

What surprised me is that $Z(s)$ seems to diverge for all values of $s$, except when $\Re(s)=\frac12$, in which case either only its real part $(\pm = +)$ or only its imaginary part $(\pm = -)$ converges. These are then equal to respectively $\Re(\zeta(\frac12 + s \, i)$ and $\Im(\zeta(\frac12 + s \, i)$ and correctly induce the non-trivial zeros as well as the zeros at $2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \phantom. \Gamma(1-s) = \pm 1$.

Question:

Can it be proven that either the real or the imaginary part of $Z(s)$ only converges for $\Re(s)=\frac12$?

Addition:

Think I can show that $\Re(Z(s))$ diverges for $\Re(s)<-1$ and $\Re(s)>2$, however can't much get closer towards $\frac12$. The solution could lie in the fact that only when $\Re(s)=\frac12$, the series $Z(s)$ is related to the real and imaginary parts of $\zeta(s)$ and in all other cases these links simply don't exist.

Note that the finite series $Z(s,v)$ can be expressed into zetas and Hurwitz zetas:

$Z(s,v):=\frac{\zeta \left( s \right) -\zeta_{H} \left(s,v+1 \right)}{2} +{ \frac {\zeta \left( s-1 \right) -\zeta_{H} \left(s-1,v+1 \right) }{1- s}}\pm \left( \frac{\zeta \left( 1-s \right) -\zeta_{H} \left(1-s,v+2 \right)}{2} -\frac {\zeta \left( -s \right) +\zeta_{H} \left(-s,v+2 \right) }{s} \right)$

For $\pm=+$ and $s=\frac12$ this becomes:

$Z(\frac12,v):= \zeta \left( \frac12 \right) -\frac{\zeta_{H} \left(\frac12,v+1 \right)+\zeta_{H} \left(\frac12,v+2 \right)}{2} - \frac{\zeta_{H} \left(-\frac12,v+1 \right) -\zeta_{H} \left(-\frac12,v+2 \right)}{\frac12}$

Since $Z(\frac12)=\zeta(\frac12)$, it follows that the other terms must converge to zero when $v \rightarrow \infty$. Under the assumption that $v$ is positive, it follows that (with help from Wolfram math and working fine):

$\displaystyle \lim_{v \to +\infty} \frac {-2\sqrt {v+1}}{4 \,v+3} \, \zeta_{H} \left(\frac12,v+1 \right)=1$

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    $\begingroup$ Such a proof would show that the so-called "critical line" is definitely a very special one. $\endgroup$ – Sylvain JULIEN Dec 14 '13 at 23:09
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    $\begingroup$ if you can do this kind of manipulation with any dirichlet series and get the same conclusion, then you won't learn anything "zeta-specific" from it $\endgroup$ – acx01b Dec 15 '13 at 13:04

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