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Is there a way to prove this?

  1. Dec 27, 2005 #1
    i would like to know how to prove this equality:

    [tex]\frac{1+2^p+3^p+.....+n^p}{\int_0^{n}dxx^p}\rightarrow{1} [/tex]

    for [tex]n\rightarrow{\infty} [/tex] of course p>0

    i don,t know if is repeated (sorry in that case)..
  2. jcsd
  3. Dec 27, 2005 #2


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    Draw the graph of y= xp. Now draw a series of rectangles, base from x= 1 to 2, height 1p= 1, base 2 to 3, height 2p, etc. Observe that each rectangle lies under the graph and so the total area of the rectangles (up to x= n) is less than the area under the curve. That is, 1 is an upper bound for the sequence as n goes to infinity. It should not be difficult to show that the sequence is increasing.
  4. Dec 27, 2005 #3


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    Go look at the integral comparison test. The method of proof is exactly what you want. (And what Integral describes)

    But, the sum is "well-known" (for integer exponents):

    [tex]1^k + 2^k + 3^k + \cdots + n^k = \frac{n^{k+1} }{k+1} + \frac{n^k}{2} + O(n^{k-1})[/tex]
    Last edited: Dec 27, 2005
  5. Dec 27, 2005 #4


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    it is easy to prove the sum ^k + ...+n^k, equals n^(k+1)/[k+1] + O(n^k), by induction.
    just use the binomial theorem on (a+1)^(k+1) - a^(k+1). Expand and add up over all a=1,...,n.
    you get a telescoping sum that shows (n+1)^(k+1) - 1^(k+1)
    = (k+1) )(1^k +....+n^k) + sums of all lower powers of integers summed from 1 to n.
    reference, p.27 of courant's calculus book vol 1, in the precalculus section.
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