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 $\log 2$.