# Log and Riemann sums

1. Mar 18, 2005

### tonix

I have been working on this problem for a while.
I am supposed to prove that $$log 2 = \lim_{n \rightarrow \infty} \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2^n}$$.

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
$$\lim_{n \rightarrow \infty} \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2^n} = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1+\frac{k}{n}}$$

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. Mar 18, 2005

### matt grime

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

3. Mar 18, 2005

### tonix

I changed the equations to latex, hope it is clear now.

4. Mar 18, 2005

### shmoe

I think you mean:

$$\log 2 = \lim_{n \rightarrow \infty} \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n}$$

Can you write log(2) in terms of an integral? How did you define the function log(x)?

5. Mar 18, 2005

### tonix

No, I (or better the author of the book) mean $$2^n$$. The problem is given just like that. log(x) is just defined the usual way.

6. Mar 19, 2005

### matt grime

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. Mar 19, 2005

### tonix

argh....
yes. I just checked the errata and it indeed is 2n and not $$2^n$$.
Well, given that it is 2n, does anybody have any ideas how to prove it using Riemann sums.

8. Mar 19, 2005

### Data

What is

$$\int_1^2 \frac{1}{x} dx$$

?

What is the Riemann sum of the above integral?

9. Mar 19, 2005

### MolotovMonarch

um, what is n?

or do you just want to know the answer without using Riemann?

10. Mar 19, 2005

### Data

for arbitrary n of course.

11. Mar 20, 2005

### tonix

ah... I get it now. Thanks.