# Computation on converging alternating zeta-like series by analytic continuation

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?

• I suppose a sign error Dec 13, 2022 at 8:30
• I think what you call alternative is actually alternating. Also the first sum is probably not over all signed integers
– lcv
Dec 13, 2022 at 9:46
• To your point, any alternating sum can be naturally split into the difference of two non-alternating sums.
– lcv
Dec 13, 2022 at 9:49
• This splitting is not valid since in a conditionally converging alternating series the order of summation matters. I guess that the assignments are chosen in such a way that the correct result occurs. Dec 13, 2022 at 11:46

$$\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} \\ \Re s>0$$
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, $$\eta(1/2)=(1-\sqrt{2})\zeta(1/2)$$