Find the sum of the infinite series 1 + cos(theta)/2 + cos(2theta)/4 + cos(3theta)/8 + ... for any real theta.

Answer and explanation

Correct answer: D. (4 - 2cos(theta)) / (5 - 4cos(theta))

  1. Step 1: Consider S = sum_{n=0}^inf (e^(itheta) / 2)^n = 1 / (1 - e^(itheta)/2) = 2 / (2 - cos(theta) - isin(theta)). Step 2. Multiply numerator and denominator by conjugate (2 - cos(theta) + isin(theta)): denominator is (2 - cos)^2 + sin^2 = 5 - 4cos(theta). Real part = (4 - 2cos(theta)) / (5 - 4cos(theta)). Answer: (4 - 2cos(theta)) / (5 - 4cos(theta)). Why not the others:* The sum of cosines represents the real part of an infinite geometric progression.

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