Computation on converging alternating zeta-like series by analytic continuation

Question

I found that to compute an converging alternating zeta-like series, I can convert it into the linear combination of zeta functions in its analytic continuation domain, e.g., I can calculate $\sum_{n\in\mathbb{Z}, n\ne 0} \frac{(-1)^n}{\sqrt{n}}$ by Hurwitz zeta function $$ \sum_{n\in\mathbb{Z}, n\ne 0} \frac{(-1)^n}{\sqrt{n}} = \sum_{n=0}^\infty \frac{2}{\sqrt{2n + 2}} - \sum_{n=0}^\infty \frac{2}{\sqrt{2n + 1}} = \sqrt{2}\left(\zeta(\frac{1}{2}, 1) - \zeta(\frac{1}{2}, \frac{1}{2})\right) $$ where $\zeta(s, a)$ is Hurwitz zeta function and $\zeta(s, a) = \sum_{n=0}^\infty \frac{1}{(n+a)^s}$ when $s > 1, a\ne 0,-1,-2,...$, and its analytic continuation elsewhere. Is this a coincidence or is there any reference that the alternating zeta function can be converted to a linear combination of zeta functions in its analytic continuation domain?

Answer 1

The "alternating zeta function" you speak of is just the Dirichlet eta function:

$$\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} \\ \Re s>0$$

It obeys the relation $$\eta(s)=(1-2^{1-s})\zeta(s)$$

And of course using the analytic continuation of $\zeta$, one can analytically continue $\eta$ to an entire function on the whole complex plane. In your example, $$\eta(1/2)=(1-\sqrt{2})\zeta(1/2)$$