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Answers for “为什么复数尾的模 |R_n(sigma + it)| 随着 sigma 的增加而减小?”

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The greatest value of ∣Rn(s)∣|R_n(s)| in the region Re⁡(s)≥1+a\operatorname{Re}(s)\ge 1+a is achieved at the boundary point s=(1+a)+0is=(1+a)+0i. This is proven by chaining two inequalities: first, the triangle inequality bounds the complex modulus by the real tail, ∣Rn(σ+it)∣≤Rn(σ)|R_n(\sigma+it)|\le R_n(\sigma); second, the term-by-term comparison for real arguments shows that Rn(σ)≤Rn(1+a)R_n(\sigma)\le R_n(1+a) when σ≥1+a\sigma\ge 1+a.

Conditions: The region is the closed half-plane Re⁡(s)≥1+a\operatorname{Re}(s)\ge 1+a with a>0a>0.; s=σ+its = \sigma + it is a complex variable.; Rn(s)=∑k=n+1∞1ksR_n(s) = \sum_{k=n+1}^{\infty} \frac{1}{k^s} is the tail sum of the zeta series.

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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.