A follow up on this MO question:
Take $s \in \mathbb{C}$, $\Re s \gt 1$, and the two infinite series:
$$Z_1(s) = \sum _{n=1}^{\infty } (-1)^n \left( \frac12 + n \right) \left( {\frac {1}{n^{s}}}-\frac {1}{(n+1)^{s}} \right)$$
and:
$$Z_2(s) = \sum _{n=1}^{\infty } (-1)^n \left( {\frac {1}{n^{s}}}-\frac {1}{(n+1)^{s}} \right)$$
By rewriting the series into Dirichlet $\eta(s)$'s:
\begin{align*} Z_1(s) &= -2 \eta(s-1) + \frac12 \\ Z_2(s) &= -2 \eta(s) + 1 \end{align*} it follows that:
$$Z_2(s-1)=Z_1(s) + \frac12$$
The non-trivial zeros of $\zeta(s)$ $+ \frac12$, now all lie (assuming RH) on the line $\Re(s)= \frac32$.
Maybe this is trivial, but the simplicity of the relation between $Z_2$ and $Z_1$ surprised me. I wondered if the term $+\frac12$ can also be derived without using the conversion into $\eta(s)$ (e.g. 'bottom up' from the individual terms of the sums)?
Thanks.
Let $(*)=2Z_2(s-1)-2Z_1(s)$, then $$(*)=\sum_{n\geqslant1}(-1)^n\left(\frac{\color{red}{2n}}{n^s}-\frac{\color{red}{2n}+2}{(n+1)^s}-(\color{red}{2n}+1)\left(\frac1{n^s}-\frac1{(n+1)^s}\right)\right). $$ The terms with $\color{red}{2n}$ in the numerator cancel each other hence one is left with $$ (*)=\sum_{n\geqslant1}(-1)^n\left(-\frac{2}{(n+1)^s}-\frac1{n^s}+\frac1{(n+1)^s}\right)=\sum_{n\geqslant1}(-1)^n\left(-\frac1{(n+1)^s}-\frac1{n^s}\right).$$ Let $a_n=(-1)^n/n^s$. Since $a_n\to0$ when $n\to\infty$, $$ (*)=\sum_{n\geqslant1}(a_{n+1}-a_n)=-a_1=+1. $$