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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
     =
 =  +
 +  +
 +  + ... +
 + ... +  .       (1)
You can write it EQUIVALENTLY in the form
.       (1)
You can write it EQUIVALENTLY in the form
     =
 =  .       (2)
This sum is the Riemann sum for the integral of the function  f(x) =
.       (2)
This sum is the Riemann sum for the integral of the function  f(x) =  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) =
  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) =  .
THEREFORE,  lim
.
THEREFORE,  lim  when n tends to infinity  is
  when n tends to infinity  is   .
.
Solved.