I found the following expansion somewhat by accident. I couldn't find a reference.
$$2^{-s}\zeta(s,a) = \sum_{n=0}^\infty (-1)^n\zeta(s,n+2a)$$
What does the general power series equal?
$$\sum_{n=0}^\infty\zeta(s,n+2a) x^n$$
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$$ \sum_{n=0}^{\infty} \zeta(s,n+2\alpha) : x^n = \frac{1}{(1-x)} \left( \zeta(s,2\alpha) - \sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+2\alpha)^s} \right)$$ – Jan 09 '25 at 00:48
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@DecarbonatedOdes this is a good answer. The sum is the lerch transcendent so it’s defined for $s<0$. – tyobrien Jan 10 '25 at 02:22