SOLUTION: Evaluate the limit of ({{{sin(pi/(3n)) + sin(2pi/(3n)) + sin(3pi/(3n))}}}+...+{{{ sin((n-2)pi/(3n)) + sin((n-1)pi/(3n))}}})/n as n goes to positive infinity (takes positive

Algebra ->  Sequences-and-series -> SOLUTION: Evaluate the limit of ({{{sin(pi/(3n)) + sin(2pi/(3n)) + sin(3pi/(3n))}}}+...+{{{ sin((n-2)pi/(3n)) + sin((n-1)pi/(3n))}}})/n as n goes to positive infinity (takes positive      Log On


   

Question 389814: Evaluate the limit of
(sin%28pi%2F%283n%29%29+%2B+sin%282pi%2F%283n%29%29+%2B+sin%283pi%2F%283n%29%29+...++sin%28%28n-2%29pi%2F%283n%29%29+%2B+sin%28%28n-1%29pi%2F%283n%29%29)/n
as n goes to positive infinity (takes positive integer values.)
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
This is equivalent to finding the average y-value of f%28x%29+=+sin+x from 0 to pi%2F3, since the x-values have equal density (the upper bound is pi%2F3 because lim%28n-%3Einfinity%2C+%28%28n-1%29pi%29%2F3n%29+=+pi%2F3)

The average y-value is found by integrating the function, then dividing by the specified range (so that the average y-value times the domain will be equal to the definite integral). The area of f%28x%29+=+sin+x from 0 to pi%2F3 is

int%28sin+x%2C+dx%2C+0%2C+pi%2F3%29+=+-cos+x from pi%2F3 to 0, which is -cos+%28pi%2F3%29+%2B+cos+0+=+1%2F2. The domain is {0, pi/3} so the average value is %281%2F2%29%2F%28pi%2F3%29+=+3%2F%282pi%29