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Question #40705

Difficulty: 8/10 | 🟢 Free

The Question

Let ω=ei2π/3\omega = e^{i2\pi/3}. Evaluate the sum S=∑n=1∞Arg((n2−ωn2−ω2)2)S = \sum_{n=1}^{\infty} \text{Arg}\left( \left( \frac{n^2-\omega}{n^2-\omega^2} \right)^2 \right) where Arg(z)∈(−π,π]\text{Arg}(z) \in (-\pi, \pi] denotes the principal argument.

Options

  • A.

    −4π3-\frac{4\pi}{3}

  • B.

    2π3\frac{2\pi}{3}

  • C.

    4π3\frac{4\pi}{3}

  • D.

    −2π3-\frac{2\pi}{3}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

V-N Method (Partial Fractions)|Roots of Unity Properties|Principal Value Branch Boundaries

Why This Question Is Tricky

Common Mistakes

This problem tests the V-N (telescopic) method integrated with complex arguments and roots of unity.

The core trap lies in the distinction between the sum of principal arguments and the principal argument of an infinite product.

Students often incorrectly assume ∑Arg(wn)=Arg(∏wn)\sum \text{Arg}(w_n) = \text{Arg}(\prod w_n).

By factoring n2−ω=(n−ω2)(n+ω2)n^2-\omega = (n-\omega^2)(n+\omega^2) and evaluating the telescopic product ∏wn→ω\prod w_n \to \omega, they arrive at 2π/32\pi/3.

However, the actual sum of the arguments monotonically accumulates to −4π/3-4\pi/3, crossing the branch cut of the complex logarithm.

Recognizing that Arg(z2)=2Arg(z)\text{Arg}(z^2) = 2\text{Arg}(z) holds here and using the inverse tangent difference formula avoids this fatal error.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterSequences and Series - Class 11
ConceptV-N Method (Partial Fractions)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree