How does the lecture use the triangle inequality to bound the modulus of the complex tail $R_n$(sigma + it)?

Conditions

• $s = \sigma + it$ where $\sigma > 1$.
• $R_n(s) = \sum_{k=n+1}^{\infty} \frac{1}{k^s}$ is the tail of the zeta series.
• The triangle inequality is applied to the infinite series.

Reasoning, step by step

1. Substitute $s=\sigma+it$ into the tail sum $R_n(s)$.
2. Rewrite the summand $\frac{1}{k^s}$ as $\frac{1}{k^{\sigma}}e^{-it(\log k)}$.
3. Apply the triangle inequality: $|\sum a_k| \le \sum |a_k|$.
4. Evaluate the modulus of the oscillatory factor: $|e^{-it(\log k)}| = 1$.
5. Conclude that $|R_n(\sigma+it)| \le \sum_{k=n+1}^{\infty} \frac{1}{k^{\sigma}} = R_n(\sigma)$.

Example

The speaker's verbal image is that the terms are vectors spiraling around rather than all pointing in the same direction, so cancellation can only make the true modulus smaller.

Common misconceptions

• Thinking that the imaginary part $t$ increases the magnitude of the terms; it only affects the phase.
• Forgetting that the triangle inequality provides an upper bound, not necessarily an equality, due to potential cancellation in the complex plane.

Watch the explanation

Explore next

• How does the real term-by-term comparison $R_n$(sigma) <= $R_n(1+a)$ help prove uniform convergence?
• Why is the maximum of |$R_n(s)$| in the half-plane Re(s) >= $1+a$ achieved at the boundary $s = 1+a$?