log, riemann sums

**slevvio** · February 13th 2008, 06:14 AM

By considering a Riemann sum of a suitable function, show that

$ln2=\lim_{x\rightarrow\infty}(\sum_{r=1}^{n}\frac{ 1}{n+r})$

This is perplexing me and any help would be appreciated, thanks.

**slevvio** · February 13th 2008, 06:53 AM

I actually just worked it out - here is my solution in case anyone is interested.

$\int_0^1f(x)dx = \lim_{n\rightarrow \infty}\frac{1}{n}f(\frac{r}{n})$

So $\frac{1}{n}f(\frac{r}{n})=\frac{1}{n+r}$

$f(\frac{r}{n})=\frac{n}{n+r}=\frac{1}{1+\frac{r}{n }}$

So $f(x)=\frac{1}{1+x}$

So $R.H.S. = \int_0^1f(x)dx=\int_0^1\frac{1}{1+x}dx=ln2 = L.H.S.$