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Evaluate the limit: lim_(n->infinity) [pi / (2n)] * (1 + cos(pi/(2n)) + cos(2*pi/(2n)) +... + cos((n-1)*pi/(2n))).

1. 0
2. 1
3. pi/2
4. 2/pi

Correct answer: 1

Solution

The expression is a Riemann sum with width pi/(2n) and sample points k*pi/(2n) for k=0,...,n-1. As n->infinity, it equals the integral of cos(x) from 0 to pi/2, which is [sin(x)] from 0 to pi/2 = 1 - 0 = 1.

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