Possible Duplicate:
Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$?
Show that
$\ln(2) = \lim_{n\rightarrow\infty}\left( \frac{1}{n + 1} + \frac{1}{n + 2} + ... + \frac{1}{2n}\right)$
by considering the lower Riemann sum of $f$ where $f(x) = \frac{1}{x}$ over $[1, 2]$
I was confused looking at the equality to begin with, since taking $n \rightarrow \infty$ for all of those terms would become $0$ right?
Anyway, I attempted it regardless.
$\sum_{k=1}^n \frac{1}{n}(f(1 + \frac{k}{n}))$
$= \sum_{k=1}^n \frac{1}{n}(\frac{1}{1+ \frac{k}{n}})$
$=\sum_{k=1}^n \frac{1}{n + k} = $ the sum from the question?
I wasn't sure what to do from here. I tried something else though:
$=\frac{1}{n + \frac{n(n+1)}{2}}$
$=\frac{2}{n^2 + 3n}$ which seemed equally useless if I'm taking $n \rightarrow \infty$ as it all becomes $0$.