Question 1184743
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Help me find the limit of ({{{1}}} + {{{sqrt(2)}}} + {{{sqrt(3)}}} + . . . + {{{sqrt(n-1)}}} + {{{sqrt(n)}}}) / n^(3/2) as n goes to +infinity. Thank you!
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<pre>
Consider the sum


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


You can write it EQUIVALENTLY in the form


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



This sum is the Riemann sum for the integral of the function  f(x) = {{{sqrt(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) = (2/3)*x^(3/2).



This difference  F(1) - F(0)  is  {{{(2/3)*1}}} - {{{(2/3)*0}}} = {{{2/3}}}.



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

Solved.