Question 1184792
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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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<pre>
Consider the sum


    {{{S[n]}}} = {{{1/(2n+1)}}} + {{{1/(2n+2)}}} + {{{1/(2n+3)}}} + ... + {{{1/(3n)}}}.       (1)


You can write it EQUIVALENTLY in the form


    {{{S[n]}}} = {{{sum((1/n)*(1/(2+k/n)), k=1, n)}}}.       (2)



This sum is the Riemann sum for the integral of the function  f(x) = {{{1/(2+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) = {{{ln(3/2)}}}.



THEREFORE,  lim {{{S[n]}}}  when n tends to infinity  is  {{{ln(3/2)}}}.
</pre>

Solved.