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Log and Riemann sums

  1. Mar 18, 2005 #1
    I have been working on this problem for a while.
    I am supposed to prove that [tex]
    log 2 = \lim_{n \rightarrow \infty} \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2^n}[/tex].

    The problem is that I have a hard time figuring out how I am supposed to prove that something is equal to a transcendental function without assuming its existence.

    First, I am supposed to let
    \lim_{n \rightarrow \infty} \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2^n} = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1+\frac{k}{n}}[/tex]

    So far so good... but then I should use Riemann sums to prove that this is equal to log 2. How can I do that?
    Last edited: Mar 18, 2005
  2. jcsd
  3. Mar 18, 2005 #2

    matt grime

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    your bracketing seems off, you mean the terms are like 1/(n+1), right.
  4. Mar 18, 2005 #3
    I changed the equations to latex, hope it is clear now.
  5. Mar 18, 2005 #4


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    I think you mean:

    \log 2 = \lim_{n \rightarrow \infty} \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n}[/tex]

    Can you write log(2) in terms of an integral? How did you define the function log(x)?
  6. Mar 18, 2005 #5
    No, I (or better the author of the book) mean [tex]2^n[/tex]. The problem is given just like that. log(x) is just defined the usual way.
  7. Mar 19, 2005 #6

    matt grime

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    The reason shmoe said 2n and not 2 to the n is that you rewriting of the series in that second form does not agree with the original one.
  8. Mar 19, 2005 #7
    yes. I just checked the errata and it indeed is 2n and not [tex]2^n[/tex].
    Well, given that it is 2n, does anybody have any ideas how to prove it using Riemann sums.
  9. Mar 19, 2005 #8
    What is

    [tex]\int_1^2 \frac{1}{x} dx[/tex]


    What is the Riemann sum of the above integral?
  10. Mar 19, 2005 #9
    um, what is n?

    or do you just want to know the answer without using Riemann?
  11. Mar 19, 2005 #10
    for arbitrary n of course.
  12. Mar 20, 2005 #11
    ah... I get it now. Thanks.
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