- #1
tonix
- 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?
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