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The Riemann zeta function can be analytically continued to $Re(s)>0$ by the infinite sum $$\zeta(s)=\frac{1}{1-2^{1-s}}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}.$$ Can we differentiate this with respect to $s$ to obtain the derivate $\zeta'(s)$ for $Re(s)>0$ ?

Assuming the sum satisfies the Cauchy-Riemann equations, I've arrived at the following:

$$\zeta'(s) = \sum_{n=1}^\infty\frac{(-1)^n2^sn^{-s}\left[\log(4)+(2^s-2)\log(n)\right]}{(2^s-2)^2},$$

but is this still valid for $Re(s)>0$ ?

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up vote 6 down vote accepted

Yes because $$\frac{1}{1-2^{1-s}}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}$$ is meromorphic on the half plane $s>0$ with a pole at $s=1$. Note that since you have a pole at $s=1$, neither formula is valid at $s=1$, so it should be the punctured half plane $s>0$ with the point $s=1$ removed.

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Taylor expansion around $x=0$ of $(1+x)^{-s}$ gives us :

$$n^{-s} - (n+1)^{-s} \sim s \; n^{-s-1}$$

thus : $$\sum n^{-s} (-1)^{n+1} = \sum (2n-1)^{-s} - (2n)^{-s} = \mathcal{O}\left(s \sum (2n)^{-s-1}\right)$$

which is an absolutely convergent Dirichlet series and thus holomorphic (the derivatives being still absolutely convergent).

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