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Integral / Sequence question

  1. Dec 29, 2005 #1
    If f is a continuous decreasing non-negative function on [1, oo) and
    [itex]a_n = \sum_{k=1}^nf(k) - \int_1^nf(t)dt[/itex].

    I need to show that a_n is a decreasing sequence of non-negative real numbers.

    I'm working on the decreasing part. If I subtract [itex]a_n - a_{n+1}[/itex] I get

    [itex]\int_1^{n+1}f(t)dt - \int_1^nf(t)dt - f(n+1)[/itex]

    but I don't know how to show this is bigger than zero. I did an example with [itex]f(x)=\frac{1}{x^2}[/itex] and of course it worked, but it didn't help me with the proof.

    Any suggestions would be appreciated.

    ETA: Let me clarify. Looking at this graphically, I realize that if I do the subtractions, I'm going to get a positive area. But I don't know how to formalize this.
     
    Last edited: Dec 29, 2005
  2. jcsd
  3. Dec 29, 2005 #2
    First, the expression for a_n - a_(n + 1) can be simplified so that it only involves a single integral. Then you can show the desired inequality by "wisely" choosing a step function for the integral.
     
  4. Dec 29, 2005 #3
    [itex]\int_1^{n+1}f(t)dt - \int_1^nf(t)dt = \int_n^{n+1}f(t)dt[/itex]
    and since f is strictly decreasing you know that this integral is going to be greater than any right end point approximation. With 1 rectangle, such an approximation is f(n+1).
     
  5. Dec 29, 2005 #4
    Damn...I'm embarrassed that I even asked.
     
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