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Consider the Riemann zeta function $$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^{s}},$$ This series converges absolutely on $\text{Re}(s)>1$. I have seen in multiple literature that this implies that $\zeta(s)$ is an analytic function on $\text{Re}(s)>1$. Nevertheless, I don't see why this is the case. Since this is stated without any arguments in the literature, my guess is that this should be a general fact about complex series or products. In the sense that a complex series, or product, that converges absolutely on a certain domain, is an analytic function on that domain. Is this true? And if so, are there any easy arguments for this? Or a reference about complex analysis that discusses this phenomenon?

Any help would be appreciated!

Math
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    Mere convergence/absolute convergence alone is not enough. However, this series converges (absolutely and) uniformly on every compact subset of ${s\in\Bbb{C},:, \text{Re}(s)>1}$ (by Weierstrass' $M$-test). It is the uniform convergence on compact subsets which preserves holomorphy/analyticity (see Morera's theorem in this regard). Henri Cartan's complex analysis text deals with convergence of sequences and series of holomorphic functions in chapter 5 (he doesn't deal with the Riemann zeta specifically, but the general theory is there). – peek-a-boo Jun 08 '22 at 19:55
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    See also How to show Riemann zeta function $\zeta(s)$ is holomorphic (except at $s=1$)? – peek-a-boo Jun 08 '22 at 19:59
  • @peek-a-boo The reason why I thought that absolute convergence would be enough, is that for the L-series of an elliptic curve, one only has absolute convergence on a certain domain. But besides not being uniformly convergent, it is still an analytic function on this domain. In which way is the analyticity obtained here? – Math Jun 08 '22 at 20:26
  • Another reason why I thought it was enough was by the sentence "The above series is a prototypical Dirichlet series that converges absolutely to an analytic function for s such that σ > 1" on wikipedia. – Math Jun 08 '22 at 20:32
  • I don't know about eliptic curves, so I won't comment on that. Regarding Wikipedia, I don't see how that sentence is (or gives the idea, or even claims to be) any form of justification for analyticity of the RIemann-zeta function on ${s\in\Bbb{C},:,\text{Re}(s)>1}$. They're just saying $\sum_{n=1}^{\infty}\frac{1}{n^s}$ converges absolutely on $\text{Re}(s)>1$, and that in this region the sum is an analytic function; but the sentence doesn't say that absolute convergence (alone) is the reason for the sum being analytic. – peek-a-boo Jun 08 '22 at 20:36
  • I am sure that you could get analyticity if you write $\frac{1}{h}(\zeta(s+h)-\zeta(s)) = \int_{\mathbb{N}} \frac{1}{h}\left(\frac{1}{n^sn^h}-\frac{1}{n^s} \right)~\mathrm{d}\Gamma(n)$ for the counting measure $\Gamma$. Then try to find a bound to use Lebesgue's theorem in order to be able to pull the limit w.r.t. $h$ into the integral. Have not tried though, yet. – Hyperbolic PDE friend Jun 11 '22 at 20:58

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Different approach (that I am not quite sure about, but should work): Let $A := \lbrace z \in \mathbb{C}: \mathrm{Re}(z)>1\rbrace$. Let $\Gamma$ be the counting measure on $2^{\mathbb{N}}$. Then, if $s \in A$ and $h \in B$ (where $B$ is a ball around $0$ such that $s+h \in A$), we have $$ \frac{1}{h}(\xi(s+h)-\xi(s)) = \sum_{n = 1}^\infty \frac{1}{h}\left(\frac{1}{n^sn^h}-\frac{1}{n^s} \right) = \int_{\mathbb{N}} \frac{1}{h}\left(\frac{1}{n^sn^h}-\frac{1}{n^s} \right)~\mathrm{d}\Gamma(n). $$ Observe: $$ \frac{1}{\lvert h \rvert}\left \lvert \frac{1}{n^sn^h}-\frac{1}{n^s} \right \rvert = \frac{1}{\lvert n^s\rvert}\left \lvert \frac{1-n^h}{hn^h} \right \rvert $$ We are interested in whether $\left\lvert \frac{1-n^h}{hn^h} \right \rvert$ is bounded on $B$. The only problem occurs when $h$ approaches $0$. Use l'Hôpital to get $$ \lim_{h \rightarrow 0} \frac{1-n^h}{hn^h} = \mathrm{Log}(N) $$ where $\mathrm{Log}$ is the logarothim's principal branch. So $\frac{1-n^h}{hn^h}$ stays bounded by some $C>0$. In total: $$ \frac{1}{\lvert h \rvert}\left \lvert \frac{1}{n^sn^h}-\frac{1}{n^s} \right \rvert \leq \frac{C}{\lvert n^s \rvert}. $$ We know that $$ \int_{\mathbb{N}} \frac{C}{\lvert n^s \rvert}~\mathrm{d}\Gamma(n) < \infty $$ and thus dominated convergence yields (limit on the inside is just the derivative) $$ \lim_{h \rightarrow 0, h \in B} \frac{1}{h}(\xi(s+h)-\xi(s)) = \sum_{n = 1}^\infty \lim_{h \rightarrow 0, h \in B} \frac{1}{h}\left(\frac{1}{n^sn^h}-\frac{1}{n^s} = \right) $$ $$ \sum_{n = 1}^\infty \frac{-\mathrm{Log(n)}}{n^s}. $$

Please correct me if you have any objections. I am not quite sure, whether my l'Hôpital argument and my derivative in the last step work.

Hyperbolic PDE friend
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