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A question about analysis

  1. Nov 21, 2012 #1
    I don't really understand why [itex]\frac{b^3}{n^3} \frac{n(n+1)(2n+1)}{6}[/itex] is close to b^3/3 if n is very large...can anybody explain this to me?

    I attached the problem...

    Thanks in advance
     

    Attached Files:

  2. jcsd
  3. Nov 21, 2012 #2
    If [itex]n[/itex] is big, then [itex]\frac{n(n+1)(2n+1)}{6} \simeq 2n^3/6[/itex] (in the sense that as [itex]n \to \infty[/itex] we have [itex]\frac{n(n+1)(2n+1)}{6} \to \frac{2n^3}{6}[/itex]).
     
  4. Nov 21, 2012 #3

    Mark44

    Staff: Mentor

    Expand n(n + 1)(2n + 1), and then factor out n3. Finally, take the limit of your expression as n gets large.
     
  5. Nov 21, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    In other words, write
    [tex] \frac{n(n+1)(2n+1)}{n^3} =
    \frac{2 n^3}{n^3} \left(1 + \frac{1}{n}\right) \left( 1 + \frac{1}{2n} \right).[/tex]

    RGV
     
  6. Nov 21, 2012 #5
    Thanks :)
     
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