Question

Cosπ/6 -(1/2)cos2π/6 + (1/3)cos3π/6 -(1/4)cos4π/6+.....(upto infinity) is equal to ? a)ln((√3-1)/√2) b) (1/2)ln(√3-1) c)-(1/2)ln(2+√3) d) ln(√2+√3)

Answer 1

# sum_{n=1}^{infty} {(-1)^n}/n cos({n pi}/6) = -1/2 log(2 + sqrt(3)) #

That's (c).

Explanation:

It's pretty clear this is negative because each group of 12 terms is one period of the cosine. Each period starts on an odd number, so negative term with the biggest fraction, and a `cos(pi/6),` pretty big at `sqrt{3}/2`. So the total for each period is negative because the negative numbers get higher weights, especially at the beginning.

There's only one negative choice, so

# sum_{n=1}^{infty} {(-1)^n}/n cos({n pi}/6) = -1/2 log(2 + sqrt(3)) #

I don't know offhand how to sum this, but Alpha does, and he agrees.