Calculate $\ln(2)$ using Riemann sum. 
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 $0$.

METHOD I
We may recall the celebre limit that yields Euler-Mascheroni constant, namely:
$$\lim_{n\to\infty} 1+\frac1{2}+\cdots+\frac{1}{n}-\ln{n}={\gamma}$$ $\tag{$\gamma$ is Euler-Mascheroni constant}$
Then everything boils down to: 
$$\lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} = \lim_{n\to\infty}{\gamma}+\ln{2n}-{\gamma}-\ln{n}= \ln{2}.$$
METHOD II
Use one of the consequences of the Lagrange's theorem applied on $\ln(x)$ function, namely:
$$\frac{1}{k+1} < \ln(k+1)-\ln(k)<\frac{1}{k} \space , \space k\in\mathbb{N} ,\space k>0$$
Taking $k=n,n+1,...,2n$ values to the inequality and then summing all relations, we get all we need in order to apply Squeeze theorem.
METHOD III
We may use Botez-Catalan identity and immediately get that:
$$\lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} = \lim_{n\to\infty} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + (-1)^{2n+1}\frac{1}{2n}= $$
$$\lim_{n\to\infty} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + (-1)^{n+1}\frac{1}{n}=\ln{2}.$$
The last series' limit is obtained by using Taylor expansion of $\ln(x+1)$ and take $x=1$
The proofs are complete.
