Find the limit of (sin(1/n) + sin(2/n) +...+sin(1))/n as n goes to infinity

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In this video we use Riemann sums to calculate the limit of the sum of sin(k/n) divided by to 1/n as n goes to infinity!
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Multipy both parts of the the fraction by 2 sin (1 / 2n). Using the formula 2sin(a)sin(b) = cos(a - b) - cos(a + b), we get a telescopic sum in the upper part of the fraction, which equals cos (1 / 2n) - cos (1 + 1 / 2n) which approaches 1 - cos(1). The lower part is 2n * sin (1 / 2n) approaches 1 so the limit of the fraction is 1 - cos(1)

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