How to Prove Log 2 Using Riemann Sums?

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Discussion Overview

The discussion revolves around proving the equality of log 2 using Riemann sums. Participants explore the formulation of the limit involving a series and its relationship to the logarithmic function, while addressing potential misunderstandings and clarifications regarding the terms involved.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the limit expression for log 2 and expresses difficulty in proving its equality without assuming the existence of the logarithmic function.
  • Another participant points out a potential error in the bracketing of terms in the limit expression.
  • Several participants suggest that the limit should be expressed with 2n instead of 2^n, prompting a discussion about the correct formulation.
  • Questions are raised about expressing log 2 in terms of an integral and the definition of the logarithmic function.
  • A participant asks about the Riemann sum corresponding to the integral of 1/x from 1 to 2, indicating a connection to the problem at hand.

Areas of Agreement / Disagreement

There is no consensus on the correct formulation of the limit expression, with some participants suggesting 2n while others initially interpreted it as 2^n. The discussion remains unresolved regarding the proof using Riemann sums.

Contextual Notes

Participants express uncertainty about the definitions and formulations used in the problem, particularly regarding the terms in the limit and the relationship to Riemann sums.

tonix
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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?
 
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your bracketing seems off, you mean the terms are like 1/(n+1), right.
 
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.
 
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)?
 
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.
 
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.
 
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.
 
What is

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

?

What is the Riemann sum of the above integral?
 
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
for arbitrary n of course.
 
  • #11
ah... I get it now. Thanks.
 

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