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