How can I solve this limit?

enter image description here

thanks you!!!


closed as off-topic by Daniel Lichtblau, MarcoB, Jens, Dr. belisarius, Sjoerd C. de Vries Oct 12 '15 at 17:57

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question cannot be answered without additional information. Questions on problems in code must describe the specific problem and include valid code to reproduce it. Any data used for programming examples should be embedded in the question or code to generate the (fake) data must be included." – Daniel Lichtblau, MarcoB, Jens, Dr. belisarius
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ How is it related to Mathematica programming? $\endgroup$ – Artes Oct 12 '15 at 11:02
  • 1
    $\begingroup$ This is not a site for homework problems that show no coding attempt. There may be related sites for that, I'm not sure. $\endgroup$ – Daniel Lichtblau Oct 12 '15 at 15:49
  • $\begingroup$ 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 mathematica.stackexchange.com/questions/95126/… $\endgroup$ – Dr. Wolfgang Hintze Oct 13 '15 at 8:55

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

   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


Not the answer you're looking for? Browse other questions tagged or ask your own question.