Why does the lecture prove uniform convergence of the zeta partial sums on Re(s) >= $1+a$ instead of on the full region Re(s) > 1?

Q: Why does the lecture prove uniform convergence of the zeta partial sums on Re(s) >= 1+a instead of on the full region Re(s) > 1?

The lecture restricts the proof to the region $\operatorname{Re}(s)\ge 1+a$ (with a fixed $a>0$) to avoid the singularity at $\operatorname{Re}(s)=1$. Near this boundary, the zeta function blows up, which prevents the partial sums from converging uniformly across the entire open half-plane $\operatorname{Re}(s)>1$. By staying a positive distance $a$ away from the boundary, the convergence becomes uniform.

The lecture restricts the proof to the region

\operatorname{Re}(s)\ge 1+a

(with a fixed

a>0

) to avoid the singularity at

\operatorname{Re}(s)=1

. Near this boundary, the zeta function blows up, which prevents the partial sums from converging uniformly across the entire open half-plane

\operatorname{Re}(s)>1

. By staying a positive distance

a

away from the boundary, the convergence becomes uniform.

Conditions

$a$ is a fixed positive constant ( $a>0$ ).
The region of interest is the closed half-plane $\operatorname{Re}(s)\ge 1+a$ .
The full defining region of the zeta series is $\operatorname{Re}(s)>1$ .

Reasoning, step by step

Identify the singularity of the zeta function at $\operatorname{Re}(s)=1$ .
Recognize that uniform convergence fails on the full region $\operatorname{Re}(s)>1$ because the error can be made arbitrarily large near the boundary.
Fix a positive constant $a>0$ to define a restricted region $\operatorname{Re}(s)\ge 1+a$ .
Prove uniform convergence on this restricted region, which stays away from the singularity.

Example

The speaker explains that if one takes the limit of the sum as $s$ approaches one, one gets infinity, so the function blows up there, and therefore chooses $a>0$ to avoid getting anywhere close to the singularity.

Common misconceptions

Assuming that pointwise convergence on an open region implies uniform convergence on that same region.
Thinking that the parameter $a$ is a variable that changes with $s$ , rather than a fixed constant defining the boundary of the region.

Watch the explanation

BilibiliRiemann zeta function Lecture 003: Analytic properties of the zeta function

0:20 – 0:54Watch this moment ↗

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Why does the lecture prove uniform convergence of the zeta partial sums on Re(s) >= 1+a1+a1+a instead of on the full region Re(s) > 1?