union of locally convergent domains regarding $\zeta(s)=\sum1/n^s$

Question

The series representation $\sum1/n^s$ of the Riemann zeta function $\zeta(s)$ is known to converge for $\sigma>1$, where $s=\sigma+it$.

In order to show an infinite series is holomorphic, we need to show the series converges uniformly.

However, the series $\sum 1/n^s$ doesn't converge uniformly in the domain $\sigma>1$. Intuitively, the closer $s$ is closer to $s=1$ the more quickly it diverges.

To address this, various sources make an argument based on locally uniform convergence.

The argument is that the series is uniformly convergent on any $\sigma>a$ where $a>1$. The argument proceeds by saying that the union of all these domains is $\sigma>1$, so the series is uniformly convergent here.

This is where I am confused. Surely $\sigma>a$ with $a>1$ means $\sigma>1$ which we agreed at the top is a domain in which the series does not converge uniformly.

Question: What am I misunderstanding about the standard argument? Where is my error?

Answer 1

As you said the series converges uniformly on $\Re(s) \ge 1+\epsilon$ (so that it is analytic there, by say Morera's theorem) and only locally uniformly on $\Re(s) >1$ (so that it is analytic there as well). Analyticity is also immediate from expanding $\zeta(z+s)=\sum_{n\ge 1} n^{-z}\sum_{k\ge 0} \frac{s^k (-\log n)^k }{k!}$ for $|s|<\Re(z)-1$, using the absolute convergence to change the order of summation, obtaining $\zeta(z+s)= \sum_{k\ge 0}s^k \sum_{n\ge 1}\frac{ n^{-z} (-\log n)^k}{k!}$

Answer 2

It is not true that the Dirichlet series for $\zeta(s)$ converges uniformly for $\mathrm{Re} s > 1$. This is also not how the "easy" arguments go to show that $\zeta(s)$ is holomorphic for $\mathrm{Re} s > 1$.

The typical argument begins as you note: for any $a > 1$, the Dirichlet series for $\zeta(s)$ converges absolutely for $\mathrm{Re} s > a$. It follows that for any compact subset $K$ contained in the halfplane $\mathrm{Re} s > 1$, the Dirichlet series for $\zeta(s)$ converges uniformly on $K$. Then Morera's theorem applies, showing that the Dirichlet series for $\zeta(s)$ is holomorphic in $\mathrm{Re} s > 1$.

Now with slightly more detail. Morera's theorem says that if $f$ is continuous in a nice domain $D$, and if for any triangle $T$ contained in $D$ we have that $$ \int_T f(z) dz = 0, $$ then $f$ is holomorphic in $D$.

Suppose now that $f_n$ is a sequence of holomorphic functions that converges uniformly to a function $f$ in every compact subset $K$ of some region $\Omega$. Then in fact $f$ is holomorphic in $\Omega$. (Sometimes stated concisely as locally uniform convergence of holomomorphic functions is holomorphic). To see this, let $D$ be a closed disk contained in $\Omega$, and let $T$ be any triangle in that disk. As each $f_n$ is holomorphic, we know that $$ \int_T f_n(z) dz = 0 \quad (\forall n). $$ As $f_n \to f$ uniformly in the closure of $D$, one can show that $$ \int_T f_n(z) dz \to \int_T f(z) dz $$ and also that $f$ is continuous. Morera's theorem now shows that $f$ is holomorphic in $D$. As this is true for any $D$ in $\Omega$, we find that $f$ is holomorphic on $\Omega$.