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 $0$.
 A: Although the terms of the sum go to $0$ as $n\to0$, there are $n$ terms and each is between $\dfrac1n$ and $\dfrac{1}{2n}$, therefore, the sum should be between $1$ and $\dfrac12$.

There at least a couple of ways to look at the sum $\displaystyle\sum_{k=1}^n\frac{1}{n+k}$:
Bound by Integrals:
Since $\frac1x$ is a decreasing function, we get that
$$
\log\left(\frac{2n+1}{n+1}\right)=\int_{1}^{n+1}\frac{\mathrm{d}x}{n+x}\le\sum_{k=1}^{n}\frac{1}{n+k}\le\int_{0}^{n}\frac{\mathrm{d}x}{n+x}=\log(2)\tag{1}
$$
Both sides of $(1)$ go to $\log(2)$ as $n\to\infty$, and the sum is sandwiched in between.
Riemann Sums:
The sum can be rewritten as
$$
\sum_{k=1}^n\frac{1}{1+\frac{k}{n}}\frac1n\tag{2}
$$
which is a Riemann sum for the integral
$$
\int_0^1\frac{1}{1+x}\mathrm{d}x\tag{3}
$$
where $\frac{k}{n}$ approximates $x$ and $\frac1n$ approximates $\mathrm{d}x$.
As $n\to\infty$, the mesh represented in $(2)$ gets finer and the sum approaches the integral, whose value is $\log(2)$.

Note that $\log(2)\,\dot{=}\,0.69314718$ is between $1$ and $\frac12$, as mentioned above.
A: You were almost there!$$\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n+k}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n}\frac{1}{1+\frac{k}{n}}=\int_0^1\frac{dx}{1+x}=\ln 2$$
The central equality above follows from the fact we know the integral on the right exists and thus it equals the limit of its Riemann sums no matter how we make those sums in the limit (as long, of course, as the interval's partition mesh tends to zero)
A: 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.
