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?

  • $\begingroup$ I suppose a sign error $\endgroup$ Dec 13, 2022 at 8:30
  • $\begingroup$ I think what you call alternative is actually alternating. Also the first sum is probably not over all signed integers $\endgroup$
    – lcv
    Dec 13, 2022 at 9:46
  • $\begingroup$ To your point, any alternating sum can be naturally split into the difference of two non-alternating sums. $\endgroup$
    – lcv
    Dec 13, 2022 at 9:49
  • $\begingroup$ 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. $\endgroup$
    – Peter
    Dec 13, 2022 at 11:46

1 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, $$\eta(1/2)=(1-\sqrt{2})\zeta(1/2)$$


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