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Answers for “尾和 R_n(s) 如何用于证明一致收敛?”

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The lecture restricts the proof to the region Re⁡(s)≥1+a\operatorname{Re}(s)\ge 1+a (with a fixed a>0a>0) to avoid the singularity at Re⁡(s)=1\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 Re⁡(s)>1\operatorname{Re}(s)>1.

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

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The tail sum, or remainder sum, Rn(s)R_n(s) is defined as the difference between the full Riemann zeta function ζ(s)\zeta(s) and its nn-th partial sum fn(s)f_n(s). Algebraically, it is the infinite series starting from k=n+1k=n+1: Rn(s)=ζ(s)−fn(s)=∑k=n+1∞1ksR_n(s) = \zeta(s) - f_n(s) = \sum_{k=n+1}^{\infty} \frac{1}{k^s}.

Conditions: ss is a complex number with Re⁡(s)>1\operatorname{Re}(s)>1.; ζ(s)=∑k=1∞1ks\zeta(s) = \sum_{k=1}^{\infty} \frac{1}{k^s} is the Riemann zeta function.; fn(s)=∑k=1n1ksf_n(s) = \sum_{k=1}^{n} \frac{1}{k^s} is the nn-th partial sum.