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

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

can be simplified a bit further into:

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

Although the two individual series diverge for $\Re(s) \le 1$, when added together they will correctly produce $\zeta(s)$ and its non-trivial zeros in the critical strip.

I wondered if a similar sum/difference of individually convergent series exists, that then still induces the non-trivial zeros and found that:

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

EDIT: or this simplified version written as a sum:

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

seem to do the trick. Both individual series appear to converge for $\Re(s)>1$, whilst also still generating the non-trivial zeros that now have been shifted up towards the line $\Re(s)=1.5$ (assuming RH), plus some additional trivial zeros at $\Re(s)=2$. This outcome can be easily explained when $Z(s)$ is rewritten as follows (note that rewriting is only valid for $\Re(s)>1$):

$$Z(s) = \dfrac{2}{1-s} \eta \left( s-1 \right)$$

with $\eta(s)=\left( 1-{2}^{1-s} \right) \zeta \left( s \right)$.


Numerical evidence from my tests with increasingly finite series (odd and even upper limits) support the convergence of the individual series for $\Re(s)>1$, however the answer to this question suggests that although convergence between $1<\Re(s) \le 2$ does occur, it is not absolute. So, is the above correct/allowed?

P.S.: Note that it doesn't really matter for the generation/location of the non-trivial zeros whether the factor used is $(1+2n)$ or $(\frac12+n)$ or $(2+4n)$ as long as it is of the form $(x+2x n)$ then the series converge and the dominant factor $\eta(s-1)$ will emerge when the difference between the series is taken.

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