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Harmonic number and natural logarithm.

  1. Jan 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Ok, basically I need to show that Ʃ 1/n (between 1 and n) (which is harmonic number) is θ (big theta) of ln(n), which means that is it bounded below and above by this function(upper and lower bound). But I don't quite understand how to prove it.


    2. Relevant equations

    I know that integral of 1/x between 1 and n is ln(n).

    3. The attempt at a solution

    I don't think that saying that the integral of 1/x between 1 and n is equal to ln(n) which is approximately equal to Ʃ 1/n (between 1 and n) is enough. But I don't know where to go from here.
    need to show:
    c1 f(n) <= Ʃ 1/n (between 1 and n) <= c1 f(n)

    where f(n) is the ln(n) and c1 and c2 are some constants. What I don't understand is how to find these constants.

    Thanks.
     
  2. jcsd
  3. Jan 22, 2013 #2

    jbunniii

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    Think of [itex]\sum 1/n[/itex] as a Riemann sum, composed of rectangles of width=1 and height=1/n. How does this compare with the integral of 1/x? Greater than? Less than? Equal?
     
  4. Jan 22, 2013 #3

    Ray Vickson

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    Use ## 1 \geq 1/x, \:x \in (1,2)##, ##1/2 \geq 1/x, \: x \in(2,3)##, etc and ##1/2 \leq 1/x, \: x \in(1,2)##, ##1/3 \leq 1/x, \: x \in (2,3)##, etc.
     
  5. Jan 22, 2013 #4
    I understand that the nth harmonic number grows about as fast as the natural logarithm of n, because the sum of ln(n) is integral of 1/x between 1 and n.

    What I don't understand is what makes ln(n) become the upper/lower bound of the nth harmonic number ? What makes it bigger or smaller, so it becomes a tight fit ?

    Thanks.
     
    Last edited: Jan 22, 2013
  6. Jan 22, 2013 #5

    haruspex

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    In each of the intervals (n-1, n), put a bound on the difference between 1/x and 1/n.
     
  7. Jan 22, 2013 #6

    haruspex

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    What is max wrt x of {1/x-1/n: n-1 < x < n}? What upper bound does that give you for ∫dx/x over that range? Summing over n, what upper bound for the integral of 1/x from 1 to n?
     
  8. Jan 22, 2013 #7

    Ray Vickson

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    Did you read my first reply? On ##x \in (1,2)## we have ## 1 \geq 1/x \geq 1/2##, and similarly for other intervals ##(k,k+1),##etc. What do you get if you integrate over x from x = 1 to x = 2 (or x = k to x = k+1)?
     
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