- #1

- 18

- 0

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?

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: