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All about sequence of functions

  1. Oct 5, 2005 #1
    Let {h_n} be a sequence of function defined on the interval (0,1) where

    h_n(x) = (n+n)x^(n-1)(1-x)

    a. find lim (n-> +oo) (integral) (from 0 to 1) h_n(x) dx.

    b. show that lim (n-> +oo) h_n(x) = 0 on (0, 1)

    c. Show that lim (n-> +oo) (integral) (from 0 to 1) h_n(x) dx is not equal to integral (from 0 to 1) (0 dx). What went wrong?

    SOlutions:

    a. lim (n-> +oo) integral (from 0 to 1) h_n(x) dx
    = lim (n-> +oo) integral (from 0 to 1) (n+n)x^(n-1)(1-x) dx
    =lim (n-> +oo)(n+n) integral (from 0 to 1)x^(n-1)(1-x) dx

    = lim (n-> +oo)(n+n) (1/n - 1/(n+1))
    = lim (n-> +oo)n(n + 1) (1/((n)(n+1))
    = 1.

    b. I used the n-th term test in proving this... because if the series of h_n(x) is convergent then lim (n-> +oo) h_n(x) = 0 on (0, 1). But by ratio test, h_n(x) is convergent because the limit of
    a(n+1)/a(n) as n -> +oo is x, but 0 < x < 1.

    c. That's the part that I got stuck... well, it seems that the statement above is true... how do I solve this?
     
  2. jcsd
  3. Oct 5, 2005 #2

    StatusX

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    Roughly speaking, the main contribution to the integral for large n comes from an increasingly small neighborhood near 1. For any given point x, no matter how close to one, hn(x) eventually gets very small as n increases. But there are still points left between x and 1 for which the value of hn evaluated at these points is very large, and this keeps the value of the integral at one. By the way, did you mean n+n^2 for the factor in front of the function?
     
  4. Oct 5, 2005 #3
    Thank you for clarifying...

    yes, it should have been n^2 + n.
     
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