Log and Riemann sums

  • Thread starter tonix
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  • #1
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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
[tex]
\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:

Answers and Replies

  • #2
matt grime
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your bracketing seems off, you mean the terms are like 1/(n+1), right.
 
  • #3
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matt grime said:
your bracketing seems off, you mean the terms are like 1/(n+1), right.

I changed the equations to latex, hope it is clear now.
 
  • #4
shmoe
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I think you mean:

[tex]
\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)?
 
  • #5
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shmoe said:
I think you mean:

[tex]
\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)?

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.
 
  • #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.
 
  • #7
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matt grime said:
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.

argh....
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.
 
  • #8
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What is

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

?

What is the Riemann sum of the above integral?
 
  • #9
Data said:
What is

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

?

What is the Riemann sum of the above integral?
um, what is n?

or do you just want to know the answer without using Riemann?
 
  • #10
998
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for arbitrary n of course.
 
  • #11
18
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ah... I get it now. Thanks.
 

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