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Harmonic number and natural logarithm.

  • Thread starter Demonoid
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  • #1
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Homework Statement



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.


Homework Equations



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

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.
 

Answers and Replies

  • #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?
 
  • #3
Ray Vickson
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Homework Statement



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.


Homework Equations



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

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.
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.
 
  • #4
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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?
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:
  • #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.
 
  • #6
haruspex
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Demonoid said:
In each of the intervals (n-1, n), put a bound on the difference between 1/x and 1/n.
I don't quite understand what you mean by that.

What I was thinking is something like this:

1/2ln(x) <=Ʃ1/n (between 1 and n) <= 5ln(x)

c is 1/2 lower bound
c is 5 upper bound

in this case Ʃ1/n (between 1 and n) is bounded by ln(x) from above and below starting from n >2.
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?
 
  • #7
Ray Vickson
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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.
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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