Where in the half-plane Re(s) >= $1+a$ does the modulus of the tail sum |$R_n(s)$| attain its greatest value?

Conditions

• The region is the closed half-plane $\operatorname{Re}(s)\ge 1+a$ with $a>0$.
• $s = \sigma + it$ is a complex variable.
• $R_n(s) = \sum_{k=n+1}^{\infty} \frac{1}{k^s}$ is the tail sum of the zeta series.

Reasoning, step by step

1. Apply the triangle inequality to get $|R_n(\sigma+it)| \le R_n(\sigma)$.
2. Use the real term-by-term comparison to get $R_n(\sigma) \le R_n(1+a)$ for $\sigma \ge 1+a$.
3. Chain the inequalities to conclude $|R_n(\sigma+it)| \le R_n(1+a)$.
4. Identify that at the boundary point $s = (1+a) + 0i$, the modulus $|R_n(s)|$ equals $R_n(1+a)$, achieving the maximum.

Example

The final slide combines the two previous estimates and adds a geometric picture. On the left, the complex plane is drawn with Real and Imaginary axes, the vertical boundary line at $1+a$, and a shaded half-plane consisting of all points with $\operatorname{Re}(s)\ge 1+a$. A sample point $\sigma+it$ is marked inside this region, with its projection $\sigma$ on the real axis. On the right, the inequalities are chained: $|R_n(\sigma+it)|\le R_n(\sigma)$ and $R_n(\sigma)\le R_n(1+a)$. By transitivity, every point in the half-plane satisfies $|R_n(s)|\le R_n(1+a)$. Since equality is attained at the boundary point $s=(1+a)+0i$, the slide concludes that the greatest value of $|R_n(s)|$ in the region $\operatorname{Re}(s)\ge 1+a$ occurs there.

Common misconceptions

• Thinking that the maximum occurs at some interior point with $t \neq 0$.
• Forgetting that taking the reciprocal reverses the inequality direction when comparing $k^{\sigma}$ and $k^{1+a}$.

Watch the explanation

Explore next

• How does the maximum of the tail sum prove uniform convergence on the region?
• Why does the real tail sum $R_n$(sigma) decrease as sigma increases?