# log, riemann sums

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• Feb 13th 2008, 07:14 AM
slevvio
log, riemann sums
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.
• Feb 13th 2008, 07:53 AM
slevvio
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 $R.H.S. = \int_0^1f(x)dx=\int_0^1\frac{1}{1+x}dx=ln2 = L.H.S.$
• Feb 16th 2008, 06:43 AM
xixihaha
thanks
yes thank you very much
i understand
btw how did u write the mathmatical symbols?
did u copy from the microsoft word???
• Feb 16th 2008, 07:00 AM
earboth
Quote:

Originally Posted by xixihaha
...
btw how did u write the mathmatical symbols?
did u copy from the microsoft word???

Hi,

have a look here: http://www.mathhelpforum.com/math-he...-tutorial.html