SOLUTION: Please help me find the limit of 1/(2n+1) + 1/(2n+2) + 1/(2n+3) + ... + 1/(3n) as n goes to +infinity. Thank you!

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Question 1184792: Please help me find the limit of 1/(2n+1) + 1/(2n+2) + 1/(2n+3) + ... + 1/(3n) as n goes to +infinity. Thank you!
Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
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Please help me find the limit of 1/(2n+1) + 1/(2n+2) + 1/(2n+3) + ... + 1/(3n) as n goes to +infinity. Thank you!
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Consider the sum


    S%5Bn%5D = 1%2F%282n%2B1%29 + 1%2F%282n%2B2%29 + 1%2F%282n%2B3%29 + ... + 1%2F%283n%29.       (1)


You can write it EQUIVALENTLY in the form


    S%5Bn%5D = sum%28%281%2Fn%29%2A%281%2F%282%2Bk%2Fn%29%29%2C+k=1%2C+n%29.       (2)



This sum is the Riemann sum for the integral of the function  f(x) = 1%2F%282%2Bx%29  over the interval [0,1].



When n tends to infinity (n---> oo), the Riemann sum (2) tends to the integral, which is equal to the difference  F(1) - F(0),

where the primitive ("antiderivative")  function  F(x)  is   F(x) = ln(2+x).



This difference  F(1) - F(0)  is  ln(3) - ln(2) = ln%283%2F2%29.



THEREFORE,  lim S%5Bn%5D  when n tends to infinity  is  ln%283%2F2%29.

Solved.