What does $\lim_{n \to \infty} \sum_{m=n}^{2n} \frac{1}{m}$ equal? 
Possible Duplicate:
Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$? 

I encountered the following sum in Boros & Moll's "Irresistible Integrals" question 5.2.11 (pg. 78):
$$\lim_{n \to \infty} \sum_{m=n}^{2n} \frac{1}{m}$$
How do I evaluate this sum to get a closed form?
 A: Another way:
$$\lim_{n\to\infty}\sum_{m=n}^{2n}\frac{1}{m}=\lim_{n\to\infty}\frac{1}{n}\sum_{m=n}^{2n}\frac{n}{m}=\int_{1/2}^1\frac{dx}{x}=\left.\log x\right|_{1/2}^1=-\log \frac{1}{2}=\log 2$$
A: If $H_x=1+\frac{1}{2}+\cdots+\frac{1}{x}$ (i.e. the $x^{th}$ Harmonic Number), we see
$$
\lim_{n \to \infty} \sum_{m=n}^{2n} \frac{1}{m}=
\lim_{n \to \infty} \frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}=
\lim_{n \to \infty}\left(1+\frac{1}{2}+\cdots + \frac{1}{2n}\right)-\left(1+\frac{1}{2}+\cdots+\frac{1}{n-1}\right)=
\lim_{n \to \infty} H_{2n}-H_{n-1}
$$
You may recall that 
$$\log x +\gamma+O(1)= H_x \quad (\text{as } x\to\infty)$$
Now, because $m \ge n \to \infty$ we may use $\log x$ instead of $H_x$.  Then we have
$$
\lim_{n \to \infty} \sum_{m=n}^{2n} \frac{1}{m}=
\lim_{n \to \infty} H_{2n}-H_{n-1}=
\lim_{n \to \infty} (\log (2n)+\gamma)-(\log (n-1)+\gamma)=
\lim_{n \to \infty} \log (2)+\log (n)-\log (n-1)=
\log (2)
$$
A quick numerical check seems to confirm this:  If $n=1000$, the sum approximately equals $0.6938972430$ - three correct decimal places of $\log 2$.
A: The Euler Maclaurin Sum Formula says that
$$
\sum_{k=1}^n\frac1k=\log(n)+\gamma+\frac1{2n}+\frac1{12n^2}+O\left(\frac1{n^4}\right)\tag{1}
$$
Therefore,
$$
\begin{align}
\sum_{k=n}^{2n}\frac1k
&=(\log(2k)+\gamma)-(\log(n-1)+\gamma)+O\left(\frac1n\right)\\
&=\log(2)+O\left(\frac1n\right)\tag{2}
\end{align}
$$
and $(2)$ shows that
$$
\lim_{n\to\infty}\sum_{k=n}^{2n}\frac1k=\log(2)\tag{3}
$$
