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

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

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

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- Feb 13th 2008, 06:14 AMslevviolog, riemann sums
By considering a Riemann sum of a suitable function, show that

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

This is perplexing me and any help would be appreciated, thanks. - Feb 13th 2008, 06:53 AMslevvio
I actually just worked it out - here is my solution in case anyone is interested.

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

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

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

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

So $\displaystyle R.H.S. = \int_0^1f(x)dx=\int_0^1\frac{1}{1+x}dx=ln2 = L.H.S.$ - Feb 16th 2008, 05:43 AMxixihahathanks
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, 06:00 AMearboth
Hi,

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