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How can I solve this limit?

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thanks you!!!

Gabri
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    How is it related to Mathematica programming? – Artes Oct 12 '15 at 11:02
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    This is not a site for homework problems that show no coding attempt. There may be related sites for that, I'm not sure. – Daniel Lichtblau Oct 12 '15 at 15:49
  • As the sum in the denominator is simply n^(5/2) the expression in question is just the Riemann sum the limit of which gives Integrate[x^(3/2),{x,0,1}] -> 2/5. The same idea was applied in my question http://mathematica.stackexchange.com/questions/95126/limit-of-partial-sums-involving-inverse-squares – Dr. Wolfgang Hintze Oct 13 '15 at 08:55

1 Answers1

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You can easily look up manual, and employ

Limit[Sum[k*Sqrt[k], {k, 1, n}]/(Sqrt[n]*Sum[2*k - 1, {k, 1, n}]), 
   n -> Infinity]

(* 2/5 *)

Also track the dependence of the limit on parameter (u) using

ListPlot[Table[{u, 
   Limit[Sum[k*Sqrt[k], {k, 1, n}]/(Sqrt[n]*Sum[u*k - 1, {k, 1, n}]), 
    n -> Infinity]}, {u, 1, 4, 0.1}]]

enter image description here

The arrow indicates value corresponding to u=2

thils
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